Given, x(2x+1)+1(2x+1)=x(2x+3)+3(2x+3) 14 ⇒ 2x^2+x+2x+1=2x^2+3x+6x+9 14 ⇒ 3x+1=9x 5 ⇒ 3x 9x= 5 1 ⇒ 6x= 6 ⇒ x= 6 6=1 ∴ x=1 ...

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Solving Equations with Variables on Both Sides

x+12x+1=x+32x...

Question

$(x + 1) (2 x + 1) = (x + 3) (2 x + 3) - 14$

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Solution

Given, $x (2 x + 1) + 1 (2 x + 1) = x (2 x + 3) + 3 (2 x + 3) - 14$

$\Rightarrow 2 x^{2} + x + 2 x + 1 = 2 x^{2} + 3 x + 6 x + 9 - 14$

$\Rightarrow 3 x + 1 = 9 x - 5$

$\Rightarrow 3 x - 9 x = - 5 - 1$

$\Rightarrow - 6 x = - 6$

$\Rightarrow x = \frac{- 6}{- 6} = 1$

$∴ x = 1$

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\leq x, t h e n l i m x \to \infty \frac{1}{n^{3}} {[1^{2} x] + [2^{2} x] + [3^{2} x] + l d o t s \dots + [n^{2} x]}

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= K \sqrt[3]{\frac{1 + x}{1 - x}} + C

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k

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