Câu hỏi của Dương Huy Vũ

Question

$\text{b)(2x+3)^2}+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2$

đề:tìm x

Nguyễn Tử Đằng · Accepted Answer

$\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2$

$\Leftrightarrow\left(2x+3\right)^2-\left(x+4\right)^2+\left(x-1\right)\left(x+1\right)+\left(x-5\right)\left(x+1\right)=5\left(x^2+4x+4\right)$

$\Leftrightarrow\left(2x+3+x+4\right)\left(2x+3-x-4\right)+\left(x+1\right)\left(x-1+x-5\right)=5x^2+20x+20$

$\Leftrightarrow\left(3x+7\right)\left(x-1\right)+\left(x+1\right)\left(2x-6\right)=5x^2+20x+20$

$\Leftrightarrow3x^2-3x+7x-7+2x^2-6x+2x-6=5x^2+20x+20$

$\Leftrightarrow5x^2-13-5x^2-20=20x$

$\Leftrightarrow-33=20x$

$\Leftrightarrow x=\dfrac{-33}{20}$

$\Rightarrow S=\left\{\dfrac{-33}{20}\right\}$Câu trả lời: $\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2$