Câu hỏi của Măm Măm

Question

Giải phương trình:

$a,\sqrt{x^2-4x+4}=\sqrt{4+2\sqrt{3}}$

$b,\sqrt{3x^2-4x}=2x-3$

Trần Thanh Phương · Accepted Answer

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

$\Leftrightarrow\left|x-2\right|=\sqrt{3}+1$

$\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{3}+1\x-2=-\sqrt{3}-1\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+3\x=-\sqrt{3}+1\end{matrix}\right.$

Vậy...

b) $\sqrt{3x^2-4x}=2x-3$  ( $x\ge\frac{2}{3}$ )

$\Leftrightarrow3x^2-4x=4x^2-12x+9$

$\Leftrightarrow x^2-8x+9=0$

$\Leftrightarrow x^2-2\cdot x\cdot4+16-7=0$

$\Leftrightarrow\left(x-4\right)^2=\left(\pm\sqrt{7}\right)^2$

$\Leftrightarrow x=4\pm\sqrt{7}$

Vậy...Câu trả lời: a) $pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}$

$\Leftrightarrow\left|x-2\right|=\sqrt{3}+1$

$\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{3}+1\x-2=-\sqrt{3}-1\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+3\x=-\sqrt{3}+1\end{matrix}\right.$

Vậy...

b) $\sqrt{3x^2-4x}=2x-3$  ( $x\ge\frac{2}{3}$ )

$\Leftrightarrow3x^2-4x=4x^2-12x+9$

$\Leftrightarrow x^2-8x+9=0$

$\Leftrightarrow x^2-2\cdot x\cdot4+16-7=0$

$\Leftrightarrow\left(x-4\right)^2=\left(\pm\sqrt{7}\right)^2$

$\Leftrightarrow x=4\pm\sqrt{7}$

Vậy...

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