Given \(\frac { 3+\sqrt { 7 } }{ 3-\sqrt { 7 } } =a+b\sqrt { 7 } \)
Multiplying the numerator and denominator by the conjugate of the denominator we get
\(\frac { (3+\sqrt { 7 } )(3+\sqrt { 7 } ) }{ (3-\sqrt { 7 } )(3+\sqrt { 7 } ) } =a+b\sqrt { 7 } \)
⇒ \(\frac { 9+7+6\sqrt { 7 } }{ { 3 }^{ 2 }-(\sqrt { 7 } )^{ 2 } } =a+b\sqrt { 7 } \)
⇒ \(\frac { 16+6\sqrt { 7 } }{ 9-7 } =a+b\sqrt { 7 } \Rightarrow \frac { 2(8+3\sqrt { 9 } ) }{ 2 } =a+b\sqrt { 7 } \)
 \(8+3\sqrt { 7 } =a+b\sqrt { 7 } \)
Comparing the like co-efficients both sides we get a = 8 and b = 3

Given log₂¹⁶ = 4
Using \({ log }_{ b }^{ m }=\frac { { log }_{ a }^{ m } }{ { log }_{ b }^{ m } } \)  we have
 \({ log }_{ 16 }^{ 2 }=\frac { { log }_{ 2 }^{ 2 } }{ { log }_{ 2 }^{ 16 } } =\frac { 1 }{ { log }_{ 2 }^{ { 2 }^{ 4 } } } \)    \([{ log }_{ a }^{ a }=1]\)
= \(\frac { 1 }{ 4{ log }_{ 2 }^{ 2 } } =\frac { 1 }{ 4 } \)    [using power rule]
∴ \({ log }_{ 16 }^{ 2 }=\frac { 1 }{ 4 } \)

Here a = 9,  b = 5, c = 0
\(\therefore\)  D = b² - 4ac = 52 - 4 (9) (0) = 25
D > 0 and it is a perfect square, the roots are real and distinct

Given roots are 7 and -3
Sum of the roots = 7+(-3) = 4
Product of the roots = 7(-3) = -21
The quadratic equation is x²- x (Sum of the roots) + Product of the roots = 0
Hence, the required quadratic equation is x²- 4x - 21 = 0

Given |4x - 5| > -2.
This means -2 < 4x - 5 < 2
⇒ -2 + 5 < 4x < 2 + 5
⇒ 3 < 4x < 7
\(⇒{3\over 4}\le x\le{7\over 4}\)
\(\therefore\) The Solution set is \(\left[ \frac { 3 }{ 4 } ,\frac { 7 }{ 4 } \right] \)

\(\left( \left[ (256)^{ \frac { -1 }{ 2 } } \right] ^{ \frac { -1 }{ 4 } } \right) ^{ 3 }\)  =  \((256)^{ \frac { -1 }{ 2 } \times \frac { -1 }{ 4 } \times 3 }\)   \([\because \frac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m-n }]\)
=  \((256)^{ \frac { 3 }{ 8 } }=({ 2 }^{ 8 })^{ \frac { 3 }{ 8 } }={ 2 }^{ 8\times \frac { 3 }{ 8 } }={ 2 }^{ 3 }=8\)

Given quadratic equation is \({ x }^{ 2 }+\sqrt { 2x } +3=0\)
\(\therefore\)  \(\alpha +\beta =\frac { -b }{ a } =-\sqrt { 2 } \)  \(\left[ \because \quad a=1,\quad b=\sqrt { 2 } ,\quad c=3\quad \Rightarrow \quad \alpha +\beta =\frac { -b }{ a } ,\quad \alpha \beta =\frac { c }{ a } \right] \)
\(\alpha \beta =\frac { 3 }{ 1 } =3\quad \)
Sum of the roots \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \beta +\alpha }{ \alpha \beta } =\frac { -\sqrt { 2 } }{ 3 } \)
Product of the roots \(\frac { 1 }{ \alpha } .\frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ 3 } \)
Hence, the required quadratic equation is x² - x (Sum of the roots) + product of the roots = 0.
\(\Rightarrow { x }^{ 2 }-x\left( \frac { -\sqrt { 2 } }{ 3 } \right) +\frac { 1 }{ 3 } =0\)
Multiplying by 3 we get,
\(3{ x }^{ 2 }+\sqrt { 2 } x+1=0\)

Let f(x) = 3x³ + 8x² + 8x + a
Since (x² + x + 1) is a factor of f(x), f(x) is divisible by x² + x + 1

Since f(x) is divisible by x² + x + 1, the remainder is zero.
∴ a - 5 = 0
⇒ a = 5

\({x^2+1\over x(x+1)^2}={A\over x}+{B\over x+1}+{c\over x(x+1)^2}\)  [Since the denominator is linear factor of repeated factors]
\(⇒\ {x^2+1\over x(x+1)^2}={A(x+)^2+Bx(x+1)+Cx\over x(x+1)^2}\)
⇒ x²+ 1 = A(x + 1)² + Bx(x + 1) + Cx
Putting x = 0, we get
1 = A ⇒ = 1
Putting x = -1, we get
1 + 1 = C(-1) ⇒ 2 = -C ⇒ C = -2
Putting x = 1 in (1) we get
2 = A(4) + B(2) + C
⇒ 24A + 2B + C   [∴ A = 1, C = -2]
⇒ 2 = 4 + 2B - 2
⇒ 2 + 2 - 4 = 2B ⇒ B =0
\(∴\ {x^2+1\over x(x+1)^2}={1\over x}+{0\over x+1}-{2\over (x+1)^2}\)
\(\frac { { x }^{ 2 }+1 }{ x\left( x+1 \right) ^{ 2 } } =\frac { 1 }{ x } \frac { 2 }{ \left( x+1 \right) ^{ 2 } } \)

Given \(\left( { x }^{ +\frac { 1 }{ 2 } }+{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ 2 } +{ 2x }^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ -\frac { 1 }{ 2 } } } =\frac { 9 }{ 2 } \)
⇒  \(x+\frac { 1 }{ x } +2=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ x } =\frac { 9 }{ 2 } -2=\frac { 9-4 }{ 2 } =\frac { 5 }{ 2 } \)
Consider \(\left( { x }^{ \frac { 1 }{ 2 } }-{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }\)
= \(x+\frac { 1 }{ x } -2x^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ \frac { 1 }{ 2 } } } \)
= \(x+\frac { 1 }{ x } -2\)    [From(1)]
=   \(\frac { 5 }{ 2 } -2=\frac { 5-4 }{ 2 } =\frac { 1 }{ 2 } \)       [using 1]
∴  \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\pm \frac { 1 }{ \sqrt { 2 } } \)
⇒ \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\frac { 1 }{ \sqrt { 2 } } \)   since x > 1

\({{7+x}\over{(1+x)(1+x^2)}}={{A}\over{1+x}}+{{Bx+C}\over{x^2+1}}\)
\(\Rightarrow\) x + 7  = A (x²+1) + (Bx + C) (x + 1)
Putting x = -1 in (1) we get,
6 = A(2) \(\Rightarrow\) A = 3
Equating the coefficient of x² in (1) we get,
0 = A + B \(\Rightarrow\) 0 = 3 + B \(\Rightarrow\) B = - 3
Putting x = 0 in (1) we get,
7 = A + C \(\Rightarrow\) 7 = 3 + C \(\Rightarrow\) C = 4
\(\therefore\)  \({{7+x}\over{(1+x)(1+x^2)}}={{A}\over{1+x}}{{Bx+C}\over{x^+1}}={{3}\over{1+x}}+\left( {{-3x+4}\over{x^2+1}} \right)\)

\({1\over x^4-1}={1\over (x^2+1)(x^2-1)}={1\over (x^2+1)(x+1)(x-1)}\)
\({1\over x^4-1}={Ax+B\over x^2+1}+{C\over x+1}+{D\over x-1}\)
\(⇒ {1\over x^2-1}={(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)\over (x^2+1)(x^2+1)}\)
⇒ 1= (Ax + B)(x + 1)(x - 1) + C(x² + 1) (x - 1) + D(x² + 1) (x + 1)
Putting x=1 in (1) we get
1 = D (2) (2) ⇒  \(D={1\over 4}\)
Putting x = -1 in we get
1 = C(2)(-2) ⇒  \(C=-{1\over 4}\)
Equating the coefficient of x³ we get
0 = A + C + D ⇒ A = - C - D
⇒ \(A={1\over 4}-{1\over 4}=0\)
⇒ A = 0
Putting x = 0 in (1) we get
1 = -B - C +D
⇒\(1=-B+{1\over 4}+{1\over 4}\)
⇒​​​​​​​ \(B=-1+{1\over 2}⇒B=-{1\over 2}\)
\(∴\ \ {1\over x^4-1}={0x-{1\over2}\over x^2+1}+{-{1\over 4}\over x+1}+{{1\over 4}\over x-1}\)
\(\Rightarrow\)  \({{1}\over{x^4-1}}={{-{{1}\over{2}}}\over{x^2+1}}-{{{{1}\over{4}}}\over{x+1}}+{{{{1}\over{4}}}\over{x-1}}=-{{1}\over{2(x^2+1)}}-{{1}\over{}4(x+1)}+{{1}\over{4(x-1)}}\)

Given inequality is \({x^3(x-1)\over x-2}>0\)
The critical numbers are 0, 1, 2
The possible intervals are (-∞, 0) (0, 1) (1, 2) (2, ∞)

The given inequality  \({x^2(x-1)\over x-2}>0\) is satisfied by the intervals (0, 1) and (2, ∞)
∴ Solution set is (0, 1)∪(2, ∞)
∴ Solution set is \((0,1)\bigcup(2, \infty)\)

Let the number of hours to complete the job is x.
Wages from the first scheme = 500 + 70x
Wages from the second scheme = 120x
Given 500 + 70x > 120x
\(\Rightarrow\)  500 > 120x -70x
\(\Rightarrow\)  500 > 50x
       
\(\Rightarrow\) 10 > x
\(\Rightarrow\) x < 10
\(\therefore\) The value of x so that the first scheme gives better wages is x = 1, 2, 3, 4, 5, 6, 7, 8, 9.

Given \(x={{\sqrt{3}-\sqrt{2}}\over{\sqrt{3}+\sqrt{2}}}\)
Multiplying by the conjugate of the denominator we get,
\(x={{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}\over{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}={{3-2\sqrt{6}+2}\over{({\sqrt{3}})^{2}-{(\sqrt{2})}^{2}}}\)
\(={{5-2\sqrt{6}}\over{3-2}}=5-2\sqrt{6}\)
Similarly, \(y={{\sqrt{3}+\sqrt{2}}\over{\sqrt{3}-\sqrt{2}}}\times{{\sqrt{3}+\sqrt{2}}\over{\sqrt{3}+\sqrt{2}}}\)
y \(={{3+2+2\sqrt{6}}\over{{(\sqrt{3})}^{2}-{(\sqrt{2})}^{2}}}={{5+2\sqrt{6}}\over{3-2}}=5+2\sqrt{6}\)
Now, x+y\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
and xy \(= (5-2\sqrt{6})(5+2\sqrt{6})=5^2-(2\sqrt{6})^2=25-4(6)=25-24=1. \)
We know that
x² + xy + y² = (x + y)² - xy = 10² - 1 = 100 - 1 = 99

LHS= \({{1}\over{3-\sqrt{8}}}-{{1}\over{\sqrt{8}-\sqrt{7}}}+{{1}\over{\sqrt{7}-\sqrt{6}}}-{{1}\over{\sqrt{6}-\sqrt{5}}}+{{1}\over{\sqrt{5}-2}}=5\)
Multiplying each term by the conjugate of the denominator we get
\(={3+\sqrt8\over (3-\sqrt8)(3+\sqrt8)}-{\sqrt8+\sqrt7\over(\sqrt8-\sqrt7)(\sqrt8+\sqrt7)}+{\sqrt7+\sqrt6\over( \sqrt7-\sqrt6)(\sqrt7+\sqrt6)}-{\sqrt6+\sqrt5\over(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}+{\sqrt5+2\over(\sqrt5-2)(\sqrt5+2)}\)
\(={3+\sqrt8\over3^2-(\sqrt8)^2}-{\sqrt8+\sqrt7\over (\sqrt8)^2-(\sqrt7)^2}+{\sqrt7+\sqrt6\over (\sqrt7)^2-( \sqrt6)^2}-{\sqrt6+\sqrt5\over(\sqrt6)^2-(\sqrt5)^2}+{\sqrt5+2\over (\sqrt5)^2-2^2}\)
\(={3+8\over 9-8}-{\sqrt8-\sqrt7\over 8-7}+{\sqrt7+\sqrt6\over 7-6}-{\sqrt6+\sqrt5\over 6-5}+{\sqrt5+2\over5-4}\)

Hence proved

x^ay^bz^c = [e^k(b-c)a.[ek^(c-a)]^b.[e^k(a-b)]^c
= e^k(b-c)a.e^k(c-a)b.e^k(a-b).c
= e^{k(ab+ac+bc+ab+ac-bc)}
= e^k(0) = e⁰ = 1
\(\Rightarrow\)   x^ay^bz^c = 1
Hence proved.

Given in equalities are \(x\le 3y, \ x\ge y\)
Suppose \(x=3y⇒{x\over 3}=y\)

If x = y

below x ≥ y is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

LHS = \(log2+16log{16\over 15}+12log{25\over 24}+7log{81\over 80}\)
\(=log2+log\left(16\over 15\right)^{16}+log\left(25\over 24\right)^{12}+log \left(81\over 80\right)^7\)
\(=log2\times{(2^4)^{16}\over (3\times5)^{16}}\times{(5^2)^{12}\over (2^2\times3)^{12}}\times{(3^4)^7\over 2^{28}\times5^7}\)
\(=log2^1\times{2^{64}\over 3^{16}}\times{5^{24}\over 2^{36}\times3^{12}}\times{3^{28}\over 2^{28}\times5^7}\)
\(=log{2^{1+64}.5^{24}.3^{28}\over 3^{16+12}.5^{16+7}.2^{36+28}}\)  \(\left[∵\ {a^m\over a^n}=a^{m-n} \right]\)

= log 2^65-64 x 5^24-23 = log 2¹ \(\times\) 5¹ = log₁₀¹⁰ = 1 = RHS

Given x =\(\sqrt { 2 } +\sqrt { 3 } \)
⇒  x³ =  \((\sqrt { 2 } +\sqrt { 3 } )​​​​^{ 2 }=2+3+2\sqrt { 6 } =5+2\sqrt { 6 } \)
∴ \(\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } =\frac { 5+2\sqrt { 6 } +1 }{ 5+2\sqrt { 6 } -2 } =\frac { 6+2\sqrt { 6 } }{ 3+2\sqrt { 6 } } \)
⇒ \(\frac { 6+2\sqrt { 6 } }{ 3+2\sqrt { 6 } } \times \frac { 3-2\sqrt { 6 } }{ 3-2\sqrt { 6 } } =\frac { (6+2\sqrt { 6 } )(3-2\sqrt { 6 } ) }{ 9-(2\sqrt { 6 } )^{ 2 } } \)
⇒ \(\frac { 18-12\sqrt { 6 } +6\sqrt { 6 } -4(\sqrt { 6 } )^{ 2 } }{ 9-24 } =\frac { 18-12\sqrt { 6 } -24 }{ -15 } \)
⇒ \(\frac { -6-6\sqrt { 3 } }{ -15 } \)
\(=\frac{2(1+\sqrt{6})}{5}=\frac{2+2 \sqrt{6}}{5}\)

Given \(\frac { 1 }{ 3-\sqrt { 8 } } -\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } -\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } -2 } \)   ..(1)
Multiplying each term by the conjugate of the denominator we get
 \(\frac { 1 }{ 3-\sqrt { 8 } } \) = \(\frac { 1 }{ 3-\sqrt { 8 } } \times \frac { 3+\sqrt { 8 } }{ 3+\sqrt { 8 } } =\frac { 3+\sqrt { 8 } }{ { 3 }^{ 2 }-\sqrt { 8 } ^{ 2 } } =\frac { 3+\sqrt { 8 } }{ 9-8 } =3+\sqrt { 8 } \)
\(\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } \)= \(\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } \times \frac { \sqrt { 8 } +\sqrt { 7 } }{ \sqrt { 8 } +\sqrt { 7 } } =\frac { \sqrt { 8 } +\sqrt { 7 } }{ 8-7 } =\frac { \sqrt { 8 } +\sqrt { 7 } }{ 1 } =\sqrt { 8 } +\sqrt { 7 } \)
\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \) =\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \times \frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } +\sqrt { 6 } } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } ^{ 2 }+\sqrt { 6 } ^{ 2 } } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ 7-6 } =\sqrt { 7 } +\sqrt { 6 } \)
\(\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } \)  = \(\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } \times \frac { \sqrt { 6 } +\sqrt { 5 } }{ \sqrt { 6 } +\sqrt { 5 } } =\frac { \sqrt { 6 } +\sqrt { 5 } }{ \sqrt { 6 } ^{ 2 }+\sqrt { 5 } ^{ 2 } } =\frac { \sqrt { 6 } +\sqrt { 5 } }{ 6-5 } \)
 \(\frac { 1 }{ \sqrt { 5 } -2 } \) =  \(\frac { 1 }{ \sqrt { 5 } -2 } \times \frac { \sqrt { 5 } +2 }{ \sqrt { 5 } +2 } =\frac { \sqrt { 5 } +2 }{ \sqrt { 5 } ^{ 2 }+2^{ 2 } } =\frac { \sqrt { 5 } +2 }{ 5-4 } =\sqrt { 5 } +2\)
Substituting all these values in (1)we get
\(\frac { 1 }{ 3-\sqrt { 8 } } -\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } -\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } -2 } \) = 5

\(\log _{a} b \log _{b} c \log _{c} a=\log _{a} c \log _{c} a=\log _{a} a=1\)

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)