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Question

Giải phương trình:

a. $\sqrt{4x+8}-\sqrt{2x-1}+\sqrt{9x+18}=0$

b. $\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0$

c. $\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5$

Học tốt · Accepted Answer

a) DK: x$\ge$-2,x$\ge$$\dfrac{1}{2}$ =>$\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0$ $\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0$ $\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0$ $\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}$ <=>25x+50=2x-1 =>23x=-51 =>x=$-\dfrac{51}{23}$(ko thỏa mãn dk) => phương trình vô nghiệm.. b) ĐKXĐ:$x\ge1,x\ge-1$ $\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0$ $\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0$ $\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\sqrt{x+1}-3=0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}x=1\x=8\end{matrix}\right.$(nhận) Vậy S={1;8} c) ĐKXĐ: $x\ge0$ $\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5$ $\Leftrightarrow-11\sqrt{2x}=-11$ $\Leftrightarrow\sqrt{2x}=1$ $\Leftrightarrow2x=1\ \Leftrightarrow x=\dfrac{1}{2}$Câu trả lời: a) DK: x$\ge$-2,x$\ge$$\dfrac{1}{2}$ =>$\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0$ $\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0$ $\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0$ $\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}$ <=>25x+50=2x-1 =>23x=-51 =>x=$-\dfrac{51}{23}$(ko thỏa mãn dk) => phương trình vô nghiệm.. b) ĐKXĐ:$x\ge1,x\ge-1$ $\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0$ $\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0$ $\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\sqrt{x+1}-3=0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}x=1\x=8\end{matrix}\right.$(nhận) Vậy S={1;8} c) ĐKXĐ: $x\ge0$ $\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5$ $\Leftrightarrow-11\sqrt{2x}=-11$ $\Leftrightarrow\sqrt{2x}=1$ $\Leftrightarrow2x=1\ \Leftrightarrow x=\dfrac{1}{2}$

DƯƠNG PHAN KHÁNH DƯƠNG · Answer

Câu a :$\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0$ ( ĐK : $x\ge\dfrac{1}{2}$ )

$\Leftrightarrow\sqrt{4x+8}+\sqrt{9x+18}=\sqrt{2x-1}$

$\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=\sqrt{2x-1}$

$\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}$

$\Leftrightarrow25\left(x+2\right)=2x-1$

$\Leftrightarrow25x+50=2x-1$

$\Leftrightarrow23x=-51$

$\Leftrightarrow x=-\dfrac{51}{23}< -\dfrac{1}{2}$

Vậy phương trình vô nghiệm .

Câu b :

$\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0$ ( ĐK : $x\ge1$ )

$\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-3\sqrt{\left(x-1\right)}=0$

$\Leftrightarrow\sqrt{\left(x-1\right)}\left(\sqrt{x+1}-3\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\sqrt{x+1}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\x=8\end{matrix}\right.$

Vậy $S=\left\{1;8\right\}$

Câu c : $\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5$ ( ĐK : $x\ge\dfrac{5}{6}$ )

$\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5$

$\Leftrightarrow-11\sqrt{2x}+11=0$

$\Leftrightarrow-11\left(\sqrt{2x}-1\right)=0$

$\Leftrightarrow\sqrt{2x}-1=0$

$\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)$

Vậy $S=\left\{\dfrac{1}{2}\right\}$

Chúc bạn học tốtCâu trả lời: Câu a :$\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0$ ( ĐK : $x\ge\dfrac{1}{2}$ )