If x - 2y = 0

2x - y = -2

x > 0, y > 0 represents the portion in the I quadrant only.
Hence, OABC is the required shaded region. 

Since the degree of the numerator is equal to the degree of the denominator, let us divide the numerator by the denominator

\(∴\ \ {x^2+x+1\over x^2-5x+6}=1+{6x-5\over x^2-5x+6 }\)
Consider \({6x-5\over x^2-5x+6}={6x-5\over (x-3)(x-2)}={A\over (x-3)}+{B\over (x-2)}\)
\(⇒\ {6x-5\over x^2-5x+6}={A(x-2)+B(x-3)\over (x-3)(x-2}\)
⇒ 6x - 5 = A(x - 2) + B(x - 3)
Putting x=2 in (2) we get
7 = B(-1) ⇒ B = -7
Putting x = 3 in (2) we get
13 = A(1) ⇒ A = 13
\(∴\ {6x-5\over x^2-5x+6}={13\over x-3}-{7\over x-2}\)
Substituting in (1) we get
\({x^2+x+1\over x^2-5x+6}=1+{13\over x-3}-{7\over x-2}\)

\({1\over x^4-1}={1\over (x^2+1)(x^2-1)}={1\over (x^2+1)(x+1)(x-1)}\)
\({1\over x^4-1}={Ax+B\over x^2+1}+{C\over x+1}+{D\over x-1}\)
\(⇒ {1\over x^2-1}={(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)\over (x^2+1)(x^2+1)}\)
⇒ 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
⇒\(1=-B+{1\over 4}+{1\over 4}\)
⇒​​​​​​​ \(B=-1+{1\over 2}⇒B=-{1\over 2}\)
\(∴\ \ {1\over x^4-1}={0x-{1\over2}\over x^2+1}+{-{1\over 4}\over x+1}+{{1\over 4}\over x-1}\)
\(\Rightarrow\)  \({{1}\over{x^4-1}}={{-{{1}\over{2}}}\over{x^2+1}}-{{{{1}\over{4}}}\over{x+1}}+{{{{1}\over{4}}}\over{x-1}}=-{{1}\over{2(x^2+1)}}-{{1}\over{}4(x+1)}+{{1}\over{4(x-1)}}\)

\({x\over (x-1)^3}={A\over x-1}+{B\over (x-1)^2}+{C\over (x-1)^3}\)
⇒ \({x\over (x-1)^3}={A\over x-1}+{B\over (x-1)^2}+{C\over (x-1)^3}\)
⇒ x = A(x - 1)²+ B(x - 1) + C
Putting x = 1 in (1) we get
1 = C
Putting x = 0 in (1) we get
0 = A - B + C
0 = A - B + 1
⇒ A - B = -1
Equating the Coefficient of x² we get
0 = A
Substituting A = 0 in (2) we get
0 - B = -1 ⇒ B = 1
\(∴\ \ {x\over (x-1)^3}={0\over x -1}+{1\over (x-1)^2}+{1\over (x-1)^3 }\)
\(={{1}\over{{(x-1)}^{2}}}+{{1}\over{{(x-1)}^{3}}}\)

Given inequality is \({{x^2-4}\over{x^2-2x-15}}\le0\)
\(⇒\ {(x+2)(x-2)\over (x-5)(x+3)}\le 0\)
The critical numbers are -2, 2, 5, -3
∴ The possible intervals are (- ∞, -3) (- 3, -2) (-2, 2)(2, 5) and (5, ∞)

The inequality \({(x+2)(x-2)\over (x-5)(x+2)}\le 0\) is satisfied by the intervals (-3, -2) and (2, 5)
∴ Solution Set is (-3, -2) \(\cup\) (2, 5)

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 x be the number of litres of 30% acid solution
∴ Total mixture = (600 + x) litres
30% of x + 12% of 600 > 5% of (600 + x)
\(⇒\ {30x\over 100}+{12\over 100}\times600>{15\over 100}(600+x)\)
⇒30x + 7200 > 9000 + 15x [Multiplying by 100]
⇒ 30x + 7200 - 15x > 9000 [Subtracting 15x]
⇒ 15x + 7200 > 900
⇒ 15x > 1800
⇒ x > 120      ....(1)
Also, 30% of x + 12% of 600 < \(18\over 100\)(600 + x)
\({30x\over 100}+{12\over 100}\times600<{18\over 100}(600 +x)\)
⇒ 30x + 7200 < 18 (600 + x) [Multiplying by 100]
⇒ 30x + 7200 < 10, 800 + 18x
⇒ 12x + 7200 < 10, 800 [Subtracting 18x]
⇒ 12x < 10, 800 - 7200 [Subtracting 7200]
⇒ 2x < 3600
⇒ \(x<{3600\over12}\)
⇒ x < 300    ....(2)
From (1) and (2), 120 < x < 300
The number of litres of the 30% acid solution will have to be greater than 120 litres and less than 300 litres.

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

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