Câu hỏi của Cao Thi Thuy Duong

Question

giải phương trình $4\sqrt{x+1}=x^2-5x+14$

Eren · Accepted Answer

ĐKXĐ: $x\ge-1$

pt $\Leftrightarrow-x+4\sqrt{x+1}-5=x^2-5x+14-5$

$\Leftrightarrow-\left(x+1-4\sqrt{x+1}+4\right)=x^2-6x+9$

$\Leftrightarrow-\left(\sqrt{x+1}-2\right)^2=\left(x-3\right)^2$

Ta thấy $VT\le0\forall x\ge-1;VP\ge0\forall x$

nên pt $\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2=0\\left(x-3\right)^2=0\end{matrix}\right.$

$\Leftrightarrow x=3\left(TM\right)$

Vậy ...Câu trả lời: ĐKXĐ: $x\ge-1$

pt $\Leftrightarrow-x+4\sqrt{x+1}-5=x^2-5x+14-5$

$\Leftrightarrow-\left(x+1-4\sqrt{x+1}+4\right)=x^2-6x+9$

$\Leftrightarrow-\left(\sqrt{x+1}-2\right)^2=\left(x-3\right)^2$

Ta thấy $VT\le0\forall x\ge-1;VP\ge0\forall x$

nên pt $\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2=0\\left(x-3\right)^2=0\end{matrix}\right.$

$\Leftrightarrow x=3\left(TM\right)$

Vậy ...