Câu hỏi của phu tran

Question

Giai phuong trinh sau

a) (2x+5)²=(x+2)²

b) x²-5x+6=0

c) 2x³+6x²=x²+3x

Hoàng Thanh Tuấn · Accepted Answer

$\left(2x+5\right)^2=\left(x+2\right)^2\Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0$$\Leftrightarrow\left(3x+7\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{3}\x=-3\end{cases}}$
$x^2-5x+6=0\Leftrightarrow x^2-6x+x-6=0\Leftrightarrow x\left(x-6\right)+\left(x-6\right)=0$$\left(x+1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=6\x=-1\end{cases}}$
$2x^3+6x^2=x^2+3x\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)$$\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\Leftrightarrow$$x=0$hoặc $x=\frac{1}{2}$hoặc $x=-3$

Câu trả lời:

$\left(2x+5\right)^2=\left(x+2\right)^2\Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0$$\Leftrightarrow\left(3x+7\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{3}\x=-3\end{cases}}$
$x^2-5x+6=0\Leftrightarrow x^2-6x+x-6=0\Leftrightarrow x\left(x-6\right)+\left(x-6\right)=0$$\left(x+1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=6\x=-1\end{cases}}$
$2x^3+6x^2=x^2+3x\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)$$\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\Leftrightarrow$$x=0$hoặc $x=\frac{1}{2}$hoặc $x=-3$