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

The means \(-{{1}\over{4}}-3\le{3\over4}x\le{1\over 4}.\)
\(⇒-{{1}\over{4}}-3≤-{3\over4}x≤{1\over4}-3\)
\(⇒-{13\over 4}≤-{3\over4}x≤-{11\over4}\)
Multiplying by 4 throughout we get,
\(-13≤-3x≤-11\)
\({-13\over -3}≥x≥{-11\over3}.\)
\(\therefore\) The Solution set is \(\left[ \frac { -11 }{ 3 } ,\frac { 13 }{ 3 } \right] \)

\(\frac { 9 }{ (x-1)(x+2)^{ 2 } } =\frac { A }{ x-1 } +\frac { B }{ x+2 } +\frac { C }{ (\quad ) } \)
= \(\frac { A(x+2)^{ 2 }+B(x-1)(x+2)+C(x-1) }{ (x-1)(x+2)(x+2)^{ 2 } } \)
Equating numerator on b/s
9=A(x+2)²+B(x-1)(x+2)+C(x-1)
put x =-2
9 = A(0)+B(0)+C(-3)
-3C = 9 ⇒ C=-3
put x =1
9 - A(1+2)² +B(0)+C(0)
9A =9
A =1
put x =0
9 =4A-2B-C
9=4(1)-2B+3
9-7 =-2B
2 = -2B
B =-1
∴ \(\frac { 9 }{ (x-1)(x+2)^{ 2 } } =\frac { 1 }{ x-1 } \frac { 1 }{ x+2 } \frac { -3 }{ (x+2)^{ 2 } } \)

Observe that a straight line can be drawn if we identify any two points on it. For example, (3, 0) and (0, 3) can be easily identified as two .points on the straight line x + y = 3.
Draw the three straight lines x + y = 3, 2x - y = 5 and -x + 2y = 3.
Now (0, 0) does not satisfy x +y ≥ 3.  Thus, the half plane bounded by x + y = 3, not containing the origin, is the solution set of x + y ≥ 3.
Similarly, the half-plane bounded by 2x - y ≤ 5 containing the origin represents the solution set of the 2x - y ≤ 5

The region represented by -x + 2y ≤ 3 is the half space bounded by the straight line the line -x + 2y = 3 that contains the origin.
The region common to the above three half planes represents the solution set of the given linear inequalities.

If 2x + 3y = 6

x + 4y = 4

x > y > 0 represents the area in the 1 quadrant.
 
All points bounded between x = 0, y = 0, x + 4y = 4 and 2x + 3y = 6 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

Given equation is x⁴+ 1
x⁴+1 = (x²)²+1²   [ \(\because\) a² + b² = (a + b)²-2b where a = x², b = 1 ]
= (x² + 1)²- 2x²(1)
= (x² + 1)²-\((-\sqrt{2}x)^2\)
= (x² + 1 + \(\sqrt{2}\) x) (x²+ 1 -\(\sqrt{2}\)x)    [ \(\because\)  a²- b² = (a + b) (a - b)]

Let x = \(\sqrt[3]{(45.4)^2\over (3.2)^2\times(6.5)^3}\)
Taking logarithms of both sides, we get
log x =  \(log\left[ (45.4)^2\over (3.2)^2\times(6.5)^3 \right]^{1/3}\)
\(={1\over 3}log\left(45.4)^2\over (3.2)^2\times(6.5)^2\right)\)
\(={1\over 3}[2log 45.4 - 2 log3.2 - 3 log 6.5]\)
\(={1\over 3}[2(1.6571) - 2(0.5052)- 3 (0.8129)]\)
\(={1\over 3}(-0.1349)= -0.0450\)
= -1 + 1 - 0.0450 = -1.9550
log x  = -1.9550
⇒ x = anti log (-1.9550)
⇒ x = 0.9016

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

The value of
\(\log _{3} 11 \log _{11} 13 \log _{13} 15 \log _{15} 27 \log _{27} 81=\log _{3} 81 \)
\(=\log _{3} 3^{4}=4 \log _{3} 3=4 \)

\(x^{2}-\mathrm{k} x+\mathrm{c}=0\)
a and b are the roots
\(\therefore a+b=k, a b=c\)
To find
\(\sqrt{(a-b)^{2}+0^{2}}=a-b=\sqrt{(a+b)^{2}-4 a b}=\sqrt{k^{2}-4 c}\)
 

\(x^{2}-k x+16=0, a^{} \& b \text { are roots }\)
\(a+b =k ; a b=16 \)
\(a^{2}+b^{2} =(a+b)^{2}-2 a b \)
\(32 =k^{2}-32 \)
\(k^{2} =64 \)
\(k =\pm 8 \Rightarrow k=-8,8 \)

\(\log _{\sqrt{x}} 0.25 =4 \)
\((\sqrt{x})^{4} =0.25 \)
\((\sqrt{x})^{4} =\frac{1}{4} \)
\(x^{2} =\frac{1}{4} \)
\(\Rightarrow x=\frac{1}{2}=0.5\)