Let\(\sqrt{7-4\sqrt{3}}=a+b\sqrt{3}\) with a,b rationals.
Squaring on both sides, we get 7-4\(\sqrt{3}\) = a²+3b²+ 2ab\(\sqrt{3}\)
So, a²+ 3b² = 7 and 2ab = -4
Therefore a = \(\frac{-2}{b}\)
From a² + 3b² = 7, we get \((\frac{-2}{b})^2+3b^2=7\), which gives \(\frac{4}{b^2}+3b^2=7\) or 3b⁴- 7b² + 4 = 0
Solving for b² we get b² = \(\frac{(7\pm\sqrt{49-48})}{8}\), which gives b² = 1 or b² = \(\frac{4}{3}.\)
Thus, b = 士1 or b = 士 \(\frac{2}{\sqrt{3}}\)
Since b is rational, we have b = 土1 and hence the corresponding values for a are ∓ 2.
Since \(\sqrt{7-4\sqrt{3}}>0\) we have \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}.\)

Let S(n) = n + (n - 1) + (n - 2) +............ + 2 + 1
= n + (n - 1) + (n - 2) +..........+ [n - (n - 2)] + [n - (n - 1)]
=\(n[1+\frac{n-1}{n}+\frac{n-2}{n}+...+\frac{n-(n-2)}{n}+\frac{n-(n-1)}{n}]\)
≤ n[1 + 1+...+1] since \(\frac{n-1}{n}<1,\frac{n-2}{n}<1,...\)
Thus, S(n) ≤ n²
Let S(n) = a + bn + cn², where a, b, c ∈ R.
Now, S(n + 1) - S(n) = n + 1
a + b(n + 1) + c(n + 1)²- [a + bn + cn²] = n + 1
b + 2cn + c = n + 1
Thus, b + c = 1 and 2c = 1 (Equating like coefficients) which give \(b={1\over2};c={1\over2}\)
Now, S(1) = 1, a + b + c = 1 which gives a = 0
Hence, S(n) = \(\frac{1}{2}n+\frac{1}{2}n^2=\frac{n(n+1)}{2},n∈N\)

reciprocal of the other
The roots are reciprocal of the other
Let α and \({1\over α}\) be the roots
\(∴\ α+{1\over α}={-b\over a}\)
and \(α.{1\overα}={c\over a}\)
\(⇒\ 1={c\over a}\)
⇒ c = a
which is the required condition

\({{7+x}\over{(1+x)(1+x^2)}}={{A}\over{1+x}}+{{Bx+C}\over{x^2+1}}\)
\(\Rightarrow\) 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
\(\therefore\)  \({{7+x}\over{(1+x)(1+x^2)}}={{A}\over{1+x}}{{Bx+C}\over{x^+1}}={{3}\over{1+x}}+\left( {{-3x+4}\over{x^2+1}} \right)\)

\({{x+12}\over{{(x+1)}^{2}(x-2)}}={{A}\over{x+1}}+{{B}\over{{(x+1)}^{2}}}+{{C}\over{x-2}}\)
\(\Rightarrow\)    x + 12 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)²  ...(1)
Putting x = - 1 in (1) we get,
11 = B(-3) \(\Rightarrow\)  \(\boxed{B={{-11}\over{}3}}\)
Putting x = 2 in (1) we get,
14C = C(9) \(\Rightarrow\)\(\boxed{C={{14}\over{9}}}\)
Equating the Co-efficient of x² in (1) we get,
0 = A + C \(\Rightarrow\) A = - C \(\Rightarrow\)  \(\boxed{A={{-14}\over{}9}}\)
\(\therefore\)  \({{x+12}\over{{(x+1)}^{2}(x-2)}}={{{{-14}\over{9}}}\over{x+1}}-{{{{11}\over{3}}}\over{{(x+1)}^{2}}}+{{{{14}\over{9}}}\over{x-2}}=-{{14}\over{9(x+1)}}-{{11}\over{3{(x+1)}^{2}}}+{{14}\over{9(x-2)}}\)

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. 

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.

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
\({x^2+x+1\over x^2-5x+6}=1+{13\over x-3}-{7\over x-2}\)

Let f(x) = ax² + bx + c be the polynomial satisfying the given conditions.
f(0) = a(0)² + b(0) + c = 1, implies that c = 1. Now the other two conditions f(-2) = 0; f(1) = 0 give 4a - 2b + c = 0 and a + b + c = 0.
Using c = 1, we get 4a - 2b = -1 and a + b = -1. Solving these two equations we get a = b = \(-\frac{1}{2}\) and thus, we have f(x) =\(-\frac{1}{2}x^2-\frac{1}{2}x+1\)

Given that \(\frac{2}{5}\) and 1+\(\sqrt{3}\) are zeros of f(x). Thus, 1-\(\sqrt{3}\)  is also a zero f(x).
Let f(x) = a\((x-\frac{2}{5})[(x-(1+\sqrt{3}))][x-(1-\sqrt{3})]=a(x-\frac{2}{5})[(x-1)^2-3]\)
Using f(0) = -8, we have, \((-\frac{2}{5}a)\)(-2) = -8 which give a = -10.
Thus the required polynomial is f(x) = (-10)(x-\(\frac{2}{5}\))[x²-2x-2] ⇒ -10x³ + 24x² + 12x - 8