Câu hỏi của Đặng Huỳnh Trâm

Question

Giúp mình nhé

Bài: Giải pt:

a) $\sqrt{X^2-9}-\sqrt{4x-12}=0$

b) $\sqrt{1-x}+\sqrt{x}=1$

c) $\sqrt{x+3}+\sqrt{x+8}=5$

d) $\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}$

TFBoys · Accepted Answer

a) $\sqrt{x^2-9}-\sqrt{4x-12}=0$ ĐK: $x\ge3$

$\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-2\sqrt{x-3}=0$

$\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\sqrt{x+3}-2=0\end{matrix}\right.$

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

Vậy x = 3

b) $\sqrt{1-x}+\sqrt{x}=1$ ĐK: $0\le x\le1$

$\Leftrightarrow1-x+x+2\sqrt{x\left(1-x\right)}=1$

$\Leftrightarrow x\left(1-x\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\end{matrix}\right.$ (Nhận)

c) $\sqrt{x+3}+\sqrt{x+8}=5$ ĐK: $x\ge-3$

Đặt $\left\{{}\begin{matrix}a=\sqrt{x+3}\ge0\b=\sqrt{x+8}\ge0\end{matrix}\right.$

$\Rightarrow\left\{{}\begin{matrix}a+b=5\b^2-a^2=5\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}a+b=5\b-a=1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}a=2\b=3\end{matrix}\right.$

$\Leftrightarrow x=1$ (Nhận)

d) $\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}$ ĐK: $-\dfrac{1}{2}\le x\le0$

$\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}=\sqrt{3x}+3\sqrt{1-2x}$

$\Leftrightarrow\sqrt{1-2x}=3\sqrt{3x}$

$\Leftrightarrow1-2x=27x$

$\Leftrightarrow x=\dfrac{1}{29}$ (Nhận)Câu trả lời: a) $\sqrt{x^2-9}-\sqrt{4x-12}=0$ ĐK: $x\ge3$