Given rational expression is 

[∵a² - b² = (a+b)(a-b)]
\(\frac { { y }^{ 2 }-15y-34 }{ { 3y }^{ 2 }-12 } \)  =  \(\frac { (y-17)(y+2) }{ 3({ y }^{ 2 }-4) } \)
= \(\frac { (y-17)(y+2) }{ 3(y-2)(y+2) } \)
=  \(\frac { y-17 }{ 3(y-2) } \)

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 a³+b³ = ab (8-3a-3b)
⇒ a³+ b³ = 8ab - 3a²b - 3ab²
⇒  a³+b³+3a²b+3ab²=8ab
⇒ (a+b)³= 2³ ab
⇒ \(\left( \frac { a+b }{ 2 } \right) ^{ 3 }\)= ab
⇒  log\(\left( \frac { a+b }{ 2 } \right) \) = log ab    [Taking logarithm on both sides]
⇒ 3 log\(\left( \frac { a+b }{ 2 } \right) \)  log a + log b   [Using power rule and products rule]
⇒  log \(\left( \frac { a+b }{ 2 } \right) \) = \(\frac { 1 }{ 3 } \) [log a + log b]

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]
∴ \({ log }_{ 16 }^{ 2 }=\frac { 1 }{ 4 } \)

Let the two distinct irrational numbers be \(\left( 2+\sqrt { 3 } \right) \) and \(\left( 4+\sqrt { 3 } \right) \).
Their difference is \(\left( 2+\sqrt { 3 } \right) -\left( 4+\sqrt { 3 } \right) =2+\sqrt { 3 } -4-\sqrt { 3 } =2-4=-2\) which is rational.

= \({ 3 }^{ -6\times \frac { 1 }{ 3 } }={ 3 }^{ 2 }=\frac { 1 }{ { 3 }^{ 2 } } =\frac { 1 }{ 9 } \)

= \(\left( { -10 }^{ 3\times \frac { -2 }{ 3 } } \right) =-{ 10 }^{ -2 }=\frac { -1 }{ { 10 }^{ 2 } } =\frac { -1 }{ 100 } \)

Required polynomial equation is
\(\mathrm{p}(x)=a x^{2}+b x+c\)
Given one root is \(1+\sqrt{5}\)
another root is \(1-\sqrt{5}\)
Required equation is s (x- a)(x- b)
\( =(x-(1+\sqrt{5}))(x-(1-\sqrt{5})) \)
\( =(x-1-\sqrt{5})(x-1+\sqrt{5}) \)
\( =(x-1)^{2}-(\sqrt{5})^{2} \)
\( =x^{2}-2 x+1-5 \)
\( =x^{2}-2 x-4\)

Let the two irrational numbers be \(5+\sqrt { 7 } \) and \(7-\sqrt { 7 } \)
Their sum = \(\left( 5+\sqrt { 7 } \right) +\left( 7-\sqrt { 7 } \right) \ = \ 5+\sqrt { 7 } +7-\sqrt { 7 } \)
= 5 + 7 = 12 which is a rational number.
Consider  the two irrational numbers \(4+\sqrt { 6 } \) and \(4-\sqrt { 6 } \)
Their product = \(\left( 4+\sqrt { 6 } \right) +\left( 4-\sqrt { 6 } \right) ={ 4 }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 }\)= 16 - 6 = 10 which is a rational number.

Let the sides of the triangle be a cm, b cm, c cm and a>b>c.
By the given conditions,
a = 3c
b = a - 2
and a + b + c > 61
From(2), b = 3c-2
Putting the values of a and b in (3) we get, 3c + (3c + 2) + c \(\ge\) 61
\(\Rightarrow\)  7c\(\ge\) 61+2
\(\Rightarrow\) 6\(\ge{{63}\over{7}}\)
\(\Rightarrow\)  c \(\ge\) 9
\(\therefore\) Minimum length of the shortest side = 9 cm.

Given inequality is \({{2x+5}\over{x-1}}>5\)
\(\Rightarrow\) \({{2x+5}\over{x-1}}-5>0\)
\(\Rightarrow\) \({{2x+5-5x+5}\over{x-1}}>0\)
\(\Rightarrow\) \({{-3x+10}\over{x-1}}>0\)
Case(i) -3x + 10 > 0, x - 1 > 0
-3x < -10, x >1
3x < 10, x >1
\(x<{{10}\over{3}},x>1\)
\(1,x<{{10}\over{3}}\)
Case(ii) -3x + 10 < 0, x - 1< 0
-3x < -10, x < 1
3x<10,  x <1
\(x>{{10}\over{3}},x<1\)
This is impossible
\(\therefore\)  The solution is \(\left(1,{{10}\over{3}} \right)\)

Suppose y = 2x

-2x + 3y = 6
-2x = 6 - 3y
\(x={6-3y\over -2}\)

All the points above the y ≥ 2x and all the points below -2x + 3y ≤ 6 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

negative of other 
Given quadratic equation is ax² + bx + c = 0
Since one root is negative of the other, let
α and -α be the roots
\(∴\ α+(α)={-b\over a}\)
\(⇒\ 0={-b\over a}=0\)
⇒ b = 0
Also α(-α) = \(c\over a\)
\(⇒\ -α^2={c\over a}\)
Hence the required condition is b = 0

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

\(\text { Let } \log _{3} \frac{1}{81}=x \Rightarrow 3^{x}=\frac{1}{81}\)
\(\Rightarrow 3^{x}=3^{-4}\)
\(\Rightarrow x=-4 \)

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

\(5 x-1 <24 \ \ \ 5 x+1 >-24 \)
\(5 x <25 \ \ 5 x >-25\)
\(x <5 \ x >-5 \)
\(x \in(-5,5)\)

\(\text { In }(-\infty, 2), \frac{|x-2|}{x-2} \text { is negative }\)
\(\operatorname{In}[2, \infty), \frac{|x-2|}{x-2} \geq 0\)