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`a)sqrt{x^2-6x+9}=2`
`<=>sqrt{(x-3)^2}=2`
`<=>|x-3|=2`
`**x-3=2`
`<=>x=5`
`**x-3=-2`
`<=>x=1`
Vậy `S={1,5}`
`b)sqrt{4x-20}+sqrt{x-5}-1/3sqrt{9x-45}=4`
đk:`x>=5`
`pt<=>2sqrt{x-5}+sqrt{x-5}-1/3*3*sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4<=>x=9`
Vậy `S={9}`
Lời giải:
a.
PT $\Leftrightarrow \sqrt{(x-3)^2}=2$
$\Leftrightarrow |x-3|=2$
$\Leftrightarrow x-3=\pm 2$
$\Leftrightarrow x=1$ hoặc $x=5$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4(x-5)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9(x-5)}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
a) \(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\Leftrightarrow\left|x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge-5\)
\(pt\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)
\(\Leftrightarrow2\sqrt{x+5}=4\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\)