X-1-2 x-1-2 x-1-x-1-2 find x

⇒x 1/2x+1+2x+1/x 1=2 ⇒(x 1)^2+(2x+1)^2/(2x+1)(x 1)=2 ⇒x^2+1 2x+4 × 2+1+4x/2x^2 2x+x 1=2 ⇒ 5x^2+2x+2 =2(2x^2 x 1) ⇒ 5x^2+2x+2 =4x^2 2x 2 ⇒ x^2+4x+4 =0 ⇒ (x+2)^2 ...

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Byju's Answer

Standard X

Mathematics

Quadratic Formula

x-1/2 x+1+2 x...

Question

$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2$ find x

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Solution

$\Rightarrow \frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2 \Rightarrow \frac{(x - 1)^{2} + (2 x + 1)^{2}}{(2 x + 1) (x - 1)} = 2 \Rightarrow \frac{x^{2} + 1 - 2 x + 4 \times 2 + 1 + 4 x}{2 x^{2} - 2 x + x - 1} = 2 \Rightarrow 5 x^{2} + 2 x + 2 = 2 (2 x^{2} - x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 4 x^{2} - 2 x - 2 \Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow (x + 2)^{2} = 0 x = - 2$

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Similar questions

$\frac{2 x - 1}{x - 2} - \frac{2 x}{x + 2} = \frac{1}{x - 2} - \frac{4}{x + 2}$ . Find the value of x.

Q. If

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 x & x^{2} & x^{3} x^{2} + 2 x & 1 & 3 x + 1 2 x & 1 - 3 x^{2} & 5 x \end{matrix} ∣ ∣ ∣ ∣ ∣

, then find

f^{'} (1)

\int \sqrt{(x^{2} + 2 x + 5)} d x

is equal to

\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}, x \neq - \frac{1}{2}, 1

\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = \frac{5}{2}

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x−12x+1+2x+1x−1=2 find x

⇒x−12x+1+2x+1x−1=2⇒(x−1)2+(2x+1)2(2x+1)(x−1)=2⇒x2+1−2x+4×2+1+4x2x2−2x+x−1=2⇒5x2+2x+2=2(2x2−x−1)⇒5x2+2x+2=4x2−2x−2⇒x2+4x+4=0⇒(x+2)2=0x=−2

$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2$ find x