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Question

undefined Giải chi tiết hộ em với ạ

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $-1\le x\le4$

$\Leftrightarrow\left(x-3\right)\sqrt{1+x}-\left(x-3\right)+x-x\sqrt{4-x}=2x^2-6x$

$\Leftrightarrow\left(x-3\right)\left(\sqrt{1+x}-1\right)+x\left(1-\sqrt{4-x}\right)=2x^2-6x$

$\Leftrightarrow\dfrac{x\left(x-3\right)}{\sqrt{1+x}+1}+\dfrac{x\left(x-3\right)}{1+\sqrt{4-x}}=2\left(x^2-3x\right)$

$\Leftrightarrow\left[{}\begin{matrix}x^2-3x=0\Rightarrow x=...\\dfrac{1}{\sqrt{1+x}+1}+\dfrac{1}{1+\sqrt{4-x}}=2\left(1\right)\end{matrix}\right.$

Xét (1), do $VT< \dfrac{1}{1}+\dfrac{1}{1}=2\Rightarrow VT< VP\Rightarrow\left(1\right)$ vô nghiệm

Vậy ...

Câu trả lời:

ĐKXĐ: $-1\le x\le4$

$\Leftrightarrow\left(x-3\right)\sqrt{1+x}-\left(x-3\right)+x-x\sqrt{4-x}=2x^2-6x$

$\Leftrightarrow\left(x-3\right)\left(\sqrt{1+x}-1\right)+x\left(1-\sqrt{4-x}\right)=2x^2-6x$

$\Leftrightarrow\dfrac{x\left(x-3\right)}{\sqrt{1+x}+1}+\dfrac{x\left(x-3\right)}{1+\sqrt{4-x}}=2\left(x^2-3x\right)$

$\Leftrightarrow\left[{}\begin{matrix}x^2-3x=0\Rightarrow x=...\\dfrac{1}{\sqrt{1+x}+1}+\dfrac{1}{1+\sqrt{4-x}}=2\left(1\right)\end{matrix}\right.$

Xét (1), do $VT< \dfrac{1}{1}+\dfrac{1}{1}=2\Rightarrow VT< VP\Rightarrow\left(1\right)$ vô nghiệm

Vậy ...