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a) \(\Leftrightarrow x^2=\sqrt{4}\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)
b) \(\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}=9\)
\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)
c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)
\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)
f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)
\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)
\(\Leftrightarrow\sqrt{2-x}=4\)
\(\Leftrightarrow2-x=16\)
hay x=-14
Câu 2:
Ta có: \(\sqrt{x^2-4x+4}=x-1\)
\(\Leftrightarrow2-x=x-1\left(x< 2\right)\)
\(\Leftrightarrow-2x=-3\)
hay \(x=\dfrac{3}{2}\left(tm\right)\)
`A=1/(x+sqrtx)+(2sqrtx)/(x-1)-1/(x-sqrtx)`
`=(sqrtx-1+2x-sqrtx-1)/(sqrtx(x-1))`
`=(2x-2)/(sqrtx(x-1))`
`=2/sqrtx`
`b)A=1`
`<=>2/sqrtx=1`
`<=>sqrtx=2`
`<=>x=4(tm)`
a: Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}-a+b=-20\\3a+b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=8-3a=8-3\cdot7=-13\end{matrix}\right.\)
\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)
C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$
Ta có: $3>2$
$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$
C2: $3\sqrt{7}=\sqrt{63}$
Ta có: $63>28$
$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$