$$\eqalign{ & {\text{According to the question,}} \cr & {\text{If ,}}2s = a + b + c \cr & \Leftrightarrow s = \frac{{a + b + c}}{2} \cr & {\text{Let, }} \cr & a = 10 \cr & b = 10 \cr & c = 10 \cr & \because s = \frac{{a + b + c}}{2} \cr & \Rightarrow s = \frac{{10 + 10 + 10}}{2} \cr & \Rightarrow s = \frac{{30}}{2} \cr & \Rightarrow s = 15 \cr & \therefore s\left( {s - c} \right) + \left( {s - a} \right)\left( {s - b} \right) \cr & = 15\left( {15 - 10} \right) + \left( {15 - 10} \right)\left( {15 - 10} \right) \cr & = 75 + 25 \cr & = 100 \cr & {\text{Now check from option, }} \cr & {\text{Option 'A' }}ab = 10 \times 10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100\left( {{\text{Satisfied}}} \right) \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 2\ , . . . . . . (i) \cr & {\text{The value of }}{x^{12}} - \frac{1}{{{x^{12}}}} =\, ? \cr & \Rightarrow {\text{If }}x = 1 \cr & \Rightarrow x + \frac{1}{x} = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & {\text{Then, }}{x^{12}} - \frac{1}{{{x^{12}}}} \cr & \Rightarrow {1^{12}} - \frac{1}{{{1^{12}}}} \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{144}}{{x \times 10}} \cr & \Rightarrow 1000 = \frac{{144}}{{10x}} \cr & \Rightarrow x = \frac{{144}}{{1000 \times 10}} \cr & \Rightarrow x = \frac{{144}}{{10000}} \cr & \Rightarrow x = 0.0144 \cr} $$

$$\eqalign{ & {x^2} = y + z \cr & {y^2} = z + x \cr & {z^2} = x + y \cr & \Rightarrow {x^2} + x = x + y + z \cr & {\text{Adding }}x{\text{ on both sides }} \cr & x\left( {x + 1} \right) = x + y + z \cr & \frac{1}{{\left( {x + 1} \right)}} = \frac{x}{{x + y + z}} \cr & {\text{Similarly,}} \cr & \frac{1}{{\left( {y + 1} \right)}} = \frac{y}{{x + y + z}} \cr & \frac{1}{{\left( {z + 1} \right)}} = \frac{z}{{x + y + z}} \cr & \therefore \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {y + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}} \cr & = \frac{x}{{x + y + z}} + \frac{y}{{x + y + z}} + \frac{z}{{x + y + z}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$

a + b + c = 0
Have values
a = 1
b = 2
c = - 3
$$ \Rightarrow \left( {\frac{{a + b}}{c} + \frac{{b + c}}{a} + \frac{{c + a}}{b}} \right)$$     $$\left( {\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}} \right)$$
$$ \Rightarrow \left( {\frac{{1 + 2}}{{ - 3}} + \frac{{2 - 3}}{1} + \frac{{ - 3 + 1}}{2}} \right)$$     $$\left( {\frac{1}{{2 - 3}} + \frac{2}{{ - 3 + 1}} + \frac{{ - 3}}{{1 + 2}}} \right)$$
$$\eqalign{ & \Rightarrow \left( { - 1 - 1 - 1} \right)\left( { - 1 - 1 - 1} \right) \cr & \Rightarrow - 3 \times - 3 \cr & \Rightarrow 9 \cr} $$

5x + 9y = 5 . . . . . (i)
125x³ + 729y³ = 120 . . . . . (ii)
From equation (i) cubing both sides
⇒ (5x + 9y)³ = 5³
⇒ 125x³ + 729y³ + 3 × 5x × 9y(5x + 9y) = 125
⇒ 125x³ + 729y³ + 135xy × 5 = 125
⇒ 120 + 135xy × 5 = 125
⇒ 135xy × 5 = 5
⇒ xy = $$\frac{1}{{135}}$$
∴ product of x & y = $$\frac{1}{{135}}$$

$$\eqalign{ & {\text{If }}2 + x\sqrt 3 = \frac{1}{{2 + \sqrt 3 }} \cr & {\text{Find }}x = ? \cr & \Rightarrow 2 + x\sqrt 3 = \frac{1}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + x \times \sqrt 3 = \frac{{2 - \sqrt 3 }}{1} \cr & \Rightarrow 2 + x\sqrt 3 = 2 - \sqrt 3 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {x^m} \times {x^n} = 1 \cr & {x^{m + n}} = {x^0}\left[ {{x^0} = 1} \right] \cr & m + n = 0 \cr & m = - n \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & {\text{Find , }}{x^7} + \frac{1}{{{x^5}}} \cr & \,\,\,\,\,\,\,x + \frac{1}{x} = 2 \cr & \Rightarrow \mathop \downarrow \limits_1 {\text{ }}\,\,\,\,{\text{ }}\mathop \downarrow \limits_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\text{Let }}x = 1 \cr & {\text{To put value in question,}} \cr & \Rightarrow {x^7} + \frac{1}{{{x^5}}} \cr & \Rightarrow {{\text{1}}^7} + \frac{1}{{{1^5}}} \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 99 \cr & \therefore {x^2} + 1 = 99x \cr & \Rightarrow 2\left( {{x^2} + 1} \right) = 2 \times 99x \cr & \Rightarrow 2{x^2} + 2 = 198x \cr & = \frac{{100x}}{{2{x^2} + 2 + 102x}} \cr & = {\text{ }}\frac{{100x}}{{198x + 102x}} \cr & = \frac{{100x}}{{300x}} \cr & = \frac{1}{3} \cr} $$

$$\eqalign{ & \frac{{4x - 3}}{x} + \frac{{4y - 3}}{y} + \frac{{4z - 3}}{z} = 0 \cr & \Rightarrow \frac{{4x}}{x} - \frac{3}{x} + \frac{{4y}}{y} - \frac{3}{y} + \frac{{4z}}{z} - \frac{3}{z} = 0 \cr & \Rightarrow 4 - \frac{3}{x} + 4 - \frac{3}{y} + 4 - \frac{3}{z} = 0 \cr & \Rightarrow 12 - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = 0 \cr & \Rightarrow - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = - 12 \cr & \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 \cr} $$

$$\eqalign{ & a = \sqrt 6 - \sqrt 5 \cr & b = \sqrt 5 - 2 \cr & c = 2 - \sqrt 3 \cr & {\text{Rationalize all the terms}} \cr & a = \frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} \times \sqrt 6 + \sqrt 5 \cr & \,\,\,\,\, = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \times \sqrt 5 + 2 \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + 2}} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} \times 2 + \sqrt 3 \cr & \,\,\,\,\,\, = \frac{1}{{2 + \sqrt 3 }} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr & {\text{Now, }} \cr & a = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr} $$
Now the term whose denominator is largest is the smallest term.
So, a < b < c

$$\eqalign{ & a:b = 2:3{\text{ and }}b:c = 4:5 \cr & a{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}\,\,\,\,\,\,\,\,{\text{ }}{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {\,\,8{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}15{\text{ }}\,\,{\text{ }}} } \cr & \therefore {a^2}:{b^2}:bc \cr & \Rightarrow {\left( 8 \right)^2}:{\left( {12} \right)^2}:\left( {12 \times 15} \right) \cr & \Rightarrow 64:144:180 \cr & \Rightarrow 16:36:45 \cr} $$

$$\eqalign{ & x = 5 - \sqrt {21} \cr & 2x = 10 - 2\sqrt {21} \,......(i) \cr & \Rightarrow 2x = {\left( {\sqrt 7 } \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2\left( {\sqrt 7 } \right)\left( {\sqrt 3 } \right) \cr & \Rightarrow 2x = {\left( {\sqrt 7 - \sqrt 3 } \right)^2} \cr & \Rightarrow x = \frac{1}{2}{\left( {\sqrt 7 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = \frac{1}{{\sqrt 2 }}\sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} \cr & \Rightarrow \sqrt x = \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 - \sqrt 3 } \right) \cr & \therefore \frac{{\sqrt x }}{{\sqrt {32 - 2x} - \sqrt {21} }} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {32 - \left( {10 - 2\sqrt {21} } \right)} - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {22 + 2\sqrt {21} } - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {{{\left( {\sqrt {21} + 1} \right)}^2}} - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {21} + 1 - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 - \sqrt 3 } \right) \cr} $$

$$\eqalign{ & {3^{x + 3}} + 7 = 250 \cr & \Rightarrow {3^{x + 3}} = 250 - 7 \cr & \Rightarrow {3^{x + 3}} = 243 \cr & \Rightarrow {3^{x + 3}} = {3^5} \cr & \Rightarrow x + 3 = 5 \cr & \Rightarrow x = 2 \cr} $$

$$\frac{1}{4} \times $$ $$\frac{2}{6} \times $$ $$\frac{3}{8} \times $$ $$\frac{4}{{10}} \times $$ $$\frac{5}{{12}} \times $$ . . . . . $$ \times \frac{{31}}{{64}}$$ $$ = \frac{1}{{{2^x}}}$$
Denominator of first term is cut by the numerator of next terms, second term to the next one and so on.
$$\eqalign{ & \Rightarrow {\left( {\frac{1}{2}} \right)^{30}} \times {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{{2^x}}} \cr & \Rightarrow {\left( {\frac{1}{2}} \right)^{30 + 6}} = \frac{1}{{{2^x}}} \cr & \Rightarrow \frac{1}{{{2^{36}}}} = \frac{1}{{{2^x}}} \cr & \Rightarrow x = 36 \cr} $$

$$\eqalign{ & a + \frac{1}{b} = 1,{\text{ }}b + \frac{1}{c} = 1 \cr & {\text{Values of }}a,{\text{ }}b,{\text{ }}c{\text{ assume}} \cr & a = \frac{1}{2} \cr & b = 2 \cr & c = - 1 \cr & \therefore abc \cr & = \frac{1}{2} \times 2 \times - 1 \cr & = - 1 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & {\text{Find, }}{x^3} + {y^3} + {z^3} = ? \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 2 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 1 - 1 \cr & \Rightarrow \left( {{x^2} + 1 - 2x} \right) + {y^2} + \left( {{z^2} + 1 - 2z} \right) = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} + {y^2} + {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} = 0 \cr & \Rightarrow x = 1 \cr & \Rightarrow {y^2} = 0 \cr & \Rightarrow y = 0 \cr & \Rightarrow {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow z = 1 \cr & {\text{Value substituted in question,}} \cr & \Rightarrow {x^3} + {y^3} + {z^3} \cr & \Rightarrow {1^3} + 0 + {1^3} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 1 \cr & {\text{Find }}\frac{2}{{{x^2} - x + 2}} = ? \cr & x + \frac{1}{x} = 1 \cr & {x^2} + 1 = x \cr & \left( {{x^2} - x} \right) = - 1 \cr & {\text{Putting value in,}} \cr & = \frac{2}{{{x^2} - x + 2}} \cr & = \frac{2}{{ - 1 + 2}} \cr & = 2 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ , }}y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & {\text{Find, }}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? \cr & \Rightarrow {\text{ }}\frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} \cr & \Rightarrow {\text{ }}\frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} = ? \cr & {\text{Now,}} \cr & {\text{ }}x + y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow x + y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} \cr & \Rightarrow {\text{ }}x + y = \frac{{2\left( {{{\sqrt 5 }^2} + {{\sqrt 3 }^2}} \right)}}{{5 - 3}} \cr & \Rightarrow x + y = 8\,.......(i) \cr & Again, \cr & x - y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}x - y = \frac{{4 \times \sqrt 5 \times \sqrt 3 }}{2} \cr & \Rightarrow x - y = 2\sqrt {15} ..............(ii) \cr & {\text{And, }}xy = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}xy = 1 \cr & {\text{Substitutes values in the question}}{\text{.}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} \cr & \Rightarrow \frac{{{8^2} - 1}}{{{{\left( {2\sqrt {15} } \right)}^2} + 1}} \cr & \Rightarrow \frac{{63}}{{61}} \cr} $$

$$\eqalign{ & 3*5 + 5*3 \cr & \Rightarrow 3*5 = 2 \times 3 - 3 \times 5 + 3 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 - 15 + 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 \cr & \Rightarrow 5*3 = 2 \times 5 - 3 \times 3 + 3 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 10 - 9 + 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16 \cr & \therefore 3*5 + 5*3 \cr & \Rightarrow 6 + 16 = 22 \cr} $$

$$\eqalign{ & \frac{{xy}}{{x + y}} = a,\,\frac{{xz}}{{x + z}} = b,\,\frac{{yz}}{{y + z}} = c{\text{ }} \cr & {\text{Now,}} \cr & \frac{{x + y}}{{xy}} = \frac{1}{a} \cr & \frac{{x + z}}{{xz}} = \frac{1}{b} \cr & \frac{{y + z}}{{yz}} = \frac{1}{c} \cr & \Rightarrow \frac{1}{y} + \frac{1}{x} = \frac{1}{a},\frac{1}{z} + \frac{1}{x} = \frac{1}{b},\frac{1}{x} + \frac{1}{y} = \frac{1}{c} \cr & {\text{Now we have to find the value of }}x \cr & \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{1}{y} + \frac{1}{x} + \frac{1}{z} + \frac{1}{x} - \frac{1}{y} - \frac{1}{z} \cr & \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{2}{x} \cr & \Rightarrow \frac{{bc + ac - ab}}{{abc}} = \frac{2}{x} \cr & \Rightarrow x = \frac{{2abc}}{{bc + ac - ab}} \cr} $$

$$\eqalign{ & xy = 8{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{So, }}\left( {x,y} \right) = \left( {1,8} \right) \cr} $$
We have to question the options and check them.
$$\eqalign{ & \left( {8,1} \right) \cr & \left( {2,4} \right) \cr & \left( {4,2} \right) \cr & \therefore {\text{ }}2x + y \cr & = 2 \times 1 + 8 \cr & = 10 \cr & 2 \times 8 + 1 = 11 \cr & 2 \times 2 + 4 = 8{\text{ }}\left( {{\text{Minimum}}} \right) \cr & 2 \times 4 + 2 = 10 \cr} $$
Hence, in this question we have all the options.
So, take all the positive factor otherwise we should have to take - ve(negative) values also.
$$\eqalign{ & \left( {x,y} \right) = \left( {1,8} \right) \cr & {\text{ }}\left( {8,1} \right) \cr & {\text{ }}\left( {2,4} \right) \cr & {\text{ }}\left( {4,2} \right) \cr & {\text{ }}\left( { - 1, - 8} \right) \cr & {\text{ }}\left( { - 8, - 1} \right) \cr & {\text{ }}\left( { - 2, - 4} \right) \cr & {\text{ }}\left( { - 4, - 2} \right) \cr} $$

$$\eqalign{ & {\left( {{x^{b + c}}} \right)^{b - c}} \times {\left( {{x^{c + a}}} \right)^{c - a}} \times {\left( {{x^{a + b}}} \right)^{a - b}}\left( {x \ne 0} \right) \cr & = {x^{{b^2} - {c^2}}} \times {x^{{c^2} - {a^2}}} \times {x^{{a^2} - {b^2}}} \cr & = {x^{{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}}} \cr & = {x^0} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{x}{a} = \frac{1}{a} - \frac{1}{x} \cr & \Rightarrow \frac{x}{a} - \frac{1}{a} = - \frac{1}{x} \cr & \Rightarrow \left( {\frac{{x - 1}}{a}} \right) = - \frac{1}{x} \cr & \Rightarrow \frac{{1 - x}}{a} = \frac{1}{x} \cr & \Rightarrow x\left( {1 - x} \right) = a \cr & \Rightarrow x - {x^2} = a \cr} $$

$$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \Rightarrow x = 2 + 1 + 2\sqrt 2 \cr & \Rightarrow x = {\left( {\sqrt 2 + 1} \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 2 + 1 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 2 - 1 \cr & \therefore \sqrt x - \frac{1}{{\sqrt x }} \cr & = \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right) \cr & = \sqrt 2 + 1 - \sqrt 2 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & \because p = 999 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {999 + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {1000} \right)}^3}} \cr & = 1000 \cr} $$

$$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$              $$ ÷ $$  $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$    $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{ & \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr & \Rightarrow a = 1 + \frac{{10}}{{101}} \cr & \Rightarrow a = \frac{{111}}{{101}} \cr & \Rightarrow b = 1 - \frac{{10}}{{101}} \cr & \Rightarrow b = \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$

$$\eqalign{ & 1 < x < 2 \cr & \sqrt {{{\left( {x - 1} \right)}^2}} {\text{ + }}\sqrt {{{\left( {3 - x} \right)}^2}} \cr & \left( {{\text{Square root cancel with square}}} \right) \cr & \therefore x - 1 + 3 - x \cr & = 2 \cr} $$

$$\eqalign{ & a*b = a + b + ab \cr & 3*4 \cr & = 3 + 4 + 3 \times 4 \cr & = 19 \cr & 2*3 \cr & = 2 + 3 + 2 \times 3 \cr & = 11 \cr & \therefore 3*4 - 2*3{\text{ }} \cr & = 19 - 11 \cr & = 8 \cr} $$