$$\eqalign{ & x + \frac{1}{x} = 5 \cr & \left( {{\text{By squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = {\left( 5 \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = 23 \cr & {\text{Now,}} \cr & \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & {\text{Divided by }}{x^2}, \cr & \Rightarrow \frac{{\frac{{{x^4}}}{{{x^2}}} + \frac{{3{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + 3x + 5 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}} + 3\left( {x + \frac{1}{x}} \right) + 5}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{23 + 3\left( 5 \right) + 5}}{{23}} \cr & \Rightarrow \frac{{43}}{{23}} \cr} $$

$$\eqalign{ & x = 3 + 2\sqrt 2 {\text{ and }}xy = 1 \cr & {y^2} = \frac{1}{{{x^2}}} \cr & y = \frac{1}{x} = \frac{1}{{3 + 2\sqrt 2 }} = 3 - 2\sqrt 2 \cr & \therefore x + \frac{1}{x} = 3 + 2\sqrt 2 + 3 - 2\sqrt 2 = 6 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 36 - 2 = 34 \cr & \frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}} \cr & = \frac{{{x^2} + \frac{1}{{{x^2}}} + 3}}{{{x^2} + \frac{1}{{{x^2}}} - 3}} \cr & = \frac{{34 + 3}}{{34 - 3}} \cr & = \frac{{37}}{{31}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & x - y = 2{\text{ and }}xy = 24 \cr & {\text{By squaring}} \cr & \Rightarrow {x^2} + {y^2} - 2xy = 4 \cr & \Rightarrow {x^2} + {y^2} - 2 \times 24 = 4 \cr & \Rightarrow {x^2} + {y^2} = 4 + 48 \cr & \Rightarrow {x^2} + {y^2} = 52 \cr} $$

$$\frac{{{x^2}}}{{{y^2}}} + tx + \frac{{{y^2}}}{4}\left( {{\text{ Given}}} \right)$$
To make it a perfect square it should be in the form
$$\eqalign{ & {{\text{A}}^2} \pm 2{\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} \pm {\text{B}}} \right)^2} \cr & = {\left( {\frac{x}{y}} \right)^2} \pm tx + {\left( {\frac{y}{2}} \right)^2} \cr & = {{\text{A}}^2} \pm 2{\text{AB}} + {{\text{B}}^2} \cr & {\text{A}} = \frac{x}{y}{\text{, B}} = \frac{y}{2}\,\,\& \,\,{\text{2AB}} = tx \cr & {\text{So, }}tx = \pm 2 \times \frac{x}{y} \times \frac{y}{2} \cr & \Rightarrow tx = \pm x \cr & \Rightarrow t = \pm 1 \cr} $$

Correct option is C.( becasue x = y)

$$\eqalign{ & \frac{{4x}}{3} + {\text{2P}} = 12 \cr & \left[ {x = 6\left( {{\text{Given}}} \right)} \right]\,\, \cr & \Rightarrow \frac{{4 \times 6}}{3} + 2{\text{P}} = 12 \cr & \Rightarrow 2{\text{P}} = 12 - 8 \cr & \Rightarrow {\text{P}} = \frac{4}{2} \cr & \Rightarrow {\text{P}} = 2 \cr} $$

Here,
Base = 3 units
Height = 4 units
Area of Δ = $$\frac{1}{2}$$ × b × h
= $$\frac{1}{2}$$ × 3 × 4
= 6 sq. units

$$\eqalign{ & {64^{x + 1}} = \frac{{64}}{{{4^x}}} \cr & \Rightarrow {\left( {{4^3}} \right)^{x + 1}} - \frac{{{4^3}}}{{{4^x}}} \cr & \Rightarrow {4^{3x + 3}} = {4^{3 - x}} \cr & \Rightarrow 3x + 3 = 3 - x \cr & \Rightarrow 4x = 0 \cr & \Rightarrow x = 0 \cr} $$

$$\eqalign{ & a{x^2} + bx + c = a{\left( {x - p} \right)^2} \cr & \Rightarrow a{x^2} + bx + c = a{\left( {{x^2} + {p^2} - 2px} \right)^2} \cr & \Rightarrow a{x^2} + bx + c = a{x^2} + a{p^2} - 2apx \cr & {\text{Comparing confficients of }}{x^2}{\text{and }}x \cr & \Rightarrow b = - 2ap \cr & \Rightarrow p = - \frac{b}{{2a}}\,.......(1) \cr & and{\text{ }}c = a{p^2} \cr & \Rightarrow c = a \times \frac{{{b^2}}}{{4{a^2}}}\left[ {{\text{From (i)}}} \right] \cr & \Rightarrow 4ac = {b^2} \cr} $$

$$\eqalign{ & p = 124 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {125} \right)}^3}} \cr & = 125 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & \left( {{\text{Assume }}x = 1{\text{, so, }}1 + 1 = 2} \right) \cr & {x^{17}}{\text{ + }}\frac{1}{{{x^{19}}}} \cr & = {\left( 1 \right)^{17}} + \frac{1}{{{{\left( 1 \right)}^{19}}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & x + \frac{1}{{4x}} = \frac{3}{2} \cr & {\text{Multiply by 2 both sides}} \cr & \therefore 2x + \frac{1}{{2x}} = 3 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2x + \frac{1}{{2x}}} \right)^3} = {\left( 3 \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3.2x.\frac{1}{{2x}}\left( {2x{\text{ + }}\frac{1}{{2x}}} \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3\left( 3 \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} = 18 \cr} $$

$$\eqalign{ & a + b = 1 \cr & {\text{By cubing,}} \cr & {a^3} + {b^3} + 3ab\left( {a + b} \right) = {1^3} \cr & \Rightarrow {a^3} + {b^3} + 3ab = 1\left[ {\because a + b = 1} \right] \cr & \Rightarrow {a^3} + {b^3} + 3ab = k \cr & {\text{From above both equations,}} \cr & k = 1 \cr} $$

$$\eqalign{ & 2x + y = 5..............(i) \cr & x + 2y = 4..............(ii) \cr & {\text{Multiply equation (ii) by 2}} \cr & 2x + 4y = 8............(iii) \cr} $$
Now subtracting equation (i) from (iii)
$$\eqalign{ & {\text{ }}2x + 4y = 8 \cr & \mathop {}\limits_ - 2x\mathop + \limits_ - \,\,y\, = 5 \cr & \overline {\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3y = 3{\text{ }}} } \cr & y = 1 \cr & x = 2 \cr & \therefore {\text{Intersection point = }}\left( {2,1} \right) \cr} $$

$$\eqalign{ & {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{ }} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = 3\left( {x + \frac{1}{x}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \,\,\,\,\left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\left[ {{{\left( {x + \frac{1}{x}} \right)}^2}} \right]^2} = {\left( 3 \right)^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} + 3 = 2\left( {a + b + c} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 3 = 2a + 2b + 2c \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} - 2b + 1 + {c^2} - 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {c - 1} \right)^2} = 0 \cr & a = 1 \cr & b = 1 \cr & c = 1 \cr & \therefore \left( {a + b + c} \right) \cr & = 1 + 1 + 1 \cr & = 3 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 5 \cr & \left[ {{\text{Squaring both sides}}} \right] \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} - 2 = 25 \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = 27 \cr} $$

a + b + c = 0
$$\frac{1}{{\left( {a + b} \right)\left( {b + c} \right)}} + $$   $$\frac{1}{{\left( {a + c} \right)\left( {b + a} \right)}} + $$   $$\frac{1}{{\left( {c + a} \right)\left( {c + b} \right)}}$$
$$\eqalign{ & \Rightarrow \frac{{\left( {a + c} \right) + \left( {b + c} \right) + \left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right)}} \cr & \Rightarrow \frac{{2\left( {a + b + c} \right)}}{{\left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right)}}{\text{ }}\left( {\because a + b + c = 0} \right) \cr & \Rightarrow 0{\text{ }} \cr} $$

$$\eqalign{ & {x^2} + {y^2} - 4x - 4y + 8 = 0 \cr & \Rightarrow {x^2} + 4 - 4x + {y^2} + 4 - 4y = 0 \cr & \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = 0 \cr & {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2 \cr & {\left( {y - 2} \right)^2} = 0 \Leftrightarrow y = 2 \cr & \therefore x - y \cr & \Rightarrow 2 - 2 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & x:y = 3:4 \cr & \therefore \frac{{5x - 2y}}{{7x + 2y}} \cr & = \frac{{5 \times 3 - 2 \times 4}}{{7 \times 3 + 2 \times 4}} \cr & = \frac{{15 - 8}}{{21 + 8}} \cr & = \frac{7}{{29}} \cr} $$

$$\eqalign{ & \frac{{2p}}{{{p^2} - 2p + 1}} = \frac{1}{4} \cr & \Rightarrow \frac{{\frac{{2p}}{p}}}{{\frac{{{p^2}}}{p} - \frac{{2p}}{p} + \frac{1}{p}}} = \frac{1}{4} \cr & \Rightarrow \frac{2}{{p + \frac{1}{p} - 2}} = \frac{1}{4} \cr & \Rightarrow p + \frac{1}{p} - 2 = 8 \cr & \Rightarrow p + \frac{1}{p} = 10 \cr} $$

$$\eqalign{ & {\text{Let , }} \cr & a = 1 \cr & b = 180 \cr & c = 1 \cr & \therefore {a^3}b = abc = 180\left( {{\text{Satisfied}}} \right) \cr & \therefore c = 1 \cr} $$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & {\text{Divide by }}x \cr & \frac{{{x^2}}}{x} + \frac{1}{x} = \frac{{3x}}{x} \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 18 \cr} $$

$$\eqalign{ & 3x + \frac{1}{{2x}} = 5 \cr & \Rightarrow {\text{Multiply both sides by }}\frac{2}{3} \cr & \therefore 3x \times \frac{2}{3} + \frac{1}{2}x \times \frac{2}{3} = 5 \times \frac{2}{3} \cr & \Rightarrow 2x + \frac{1}{{3x}} = \frac{{10}}{3} \cr & \therefore {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 3.2x.\frac{1}{{3x}}\left( {{\text{ 2}}x{\text{ + }}\frac{1}{{3x}}} \right) = {\left( {\frac{{10}}{3}} \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 2\left( {\frac{{10}}{3}} \right) = \left( {\frac{{1000}}{{27}}} \right) \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000}}{{27}} - \frac{{20}}{3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000 - 180}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{820}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = 30\frac{{10}}{{27}} \cr} $$

$$\eqalign{ & x + y = z \cr & x + y - z = 0 \cr & {\text{If }}a + b + c = 0 \cr & {\text{then }}{a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \Rightarrow {x^3} + {y^3} - {z^3} = - 3xyz \cr & \therefore {x^3} + {y^3} - {z^3} + 3xyz = 0 \cr & \Rightarrow 3xyz - 3xyz = 0 \cr} $$

$$\eqalign{ & {a^3} - {b^3} - {c^3} - 3abc = 0 \cr & \therefore a - b - c = 0 \cr & \Rightarrow a = b + c \cr} $$

$$\eqalign{ & n = 7 + 4\sqrt 3 \cr & \Rightarrow n = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow n = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} + 2 \times 2 \times \sqrt 3 \cr & \Rightarrow n = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt n = 2 + \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt n }} = 2 - \sqrt 3 \cr & \therefore \sqrt n + \frac{1}{{\sqrt n }} \cr & = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & = 4 \cr} $$

x = b + c - 2a
y = c + a - 2b
z = a + b - 2c
⇒ x + y + z = (b + c - 2a) + (c + a - 2b) + (a + b - 2c) = 0
∴ Now,
⇒ x² + y² + 2xy - z²
⇒ (x + y)² - z² [∴ (a² - b²) = (a + b) (a - b)]
⇒ (x + y - z) (x + y + z)
As we know, (x + y + z) = 0
∴ x² + y² - z² + 2xy
= 0 × (x + y - z)
= 0

$$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & \left( {{\text{Take cube on both sides}}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \therefore {x^6} = - 1 \cr & \therefore {x^{18}} + {x^{12}} + {x^6} + 1 \cr & = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 \cr & = - 1 + 1 - 1 + 1 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{ }}{a^{\frac{1}{3}}} = 11 \cr & \Leftrightarrow a = {11^3} = 1331 \cr & {a^2} - 331a \cr & = a\left( {a - 331} \right) \cr & = 1331\left( {1331 - 331} \right) \cr & = 1331\left( {1000} \right) \cr & = 1331000 \cr} $$