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a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)
\(\Leftrightarrow\sqrt{x+3}=2\)
\(\Leftrightarrow x+3=4\)
\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )
c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )
\(\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)
\(\Leftrightarrow2\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )
Vậy.......
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Giải:
\(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
ĐKXĐ: \(x-2\ge0\Leftrightarrow x\ge2\)
\(7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow2\sqrt{x-2}=8\)
\(\Leftrightarrow\sqrt{x-2}=4\)
\(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa mãn)
Vậy ...
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a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
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a: \(=2\sqrt{x-3}+3\sqrt{x-3}-4\sqrt{x-3}+3-x\)
\(=\sqrt{x-3}+3-x\)
c: \(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=18\)
=>2 căn x-2=18
=>x-2=81
=>x=83
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a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
Giải phương trình:
a) \(2\sqrt{4x-8}-\dfrac{2}{3}\sqrt{9x-18}=\sqrt{49x-98}-10\)
b) \(x-\sqrt{x-1}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\ge2\\ PT\Leftrightarrow4\sqrt{x-2}-2\sqrt{x-2}-7\sqrt{x-2}=-10\\ \Leftrightarrow-5\sqrt{x-2}=-10\\ \Leftrightarrow\sqrt{x-2}=2\Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\\ b,ĐK:x\ge1\\ PT\Leftrightarrow x-3=\sqrt{x-1}\\ \Leftrightarrow x^2-6x+9=x-1\\ \Leftrightarrow x^2-7x+10=0\\ \Leftrightarrow\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge2\)
Từ pt đã cho suy ra:
\(7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
⇒ \(2\sqrt{x-2}=8\) ⇒ \(x=18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;-1\right\}\)
\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)
\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)
Vậy \(S=\left\{28\right\}\)
Đk: x >/ 5
pt đã cho \(\Leftrightarrow7\sqrt{x-2}-14\cdot\dfrac{\sqrt{x-2}}{7}-3\sqrt{x-5}=4\)
\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-5}=4\)
\(\Leftrightarrow5\sqrt{x-2}-3\sqrt{x-5}=4\)
\(\Leftrightarrow5\sqrt{x-2}=4+3\sqrt{x-5}\)
\(\Leftrightarrow25x-50=16+9x-45+24\sqrt{x-5}\)
\(\Leftrightarrow16x-21=24\sqrt{x-5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-672x+441=576x-2880\\x\ge\dfrac{21}{16}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-1248x+3321=0\\x\ge\dfrac{21}{16}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-1248x+3321=0\left(vn\right)\\x\ge\dfrac{21}{16}\end{matrix}\right.\)
Kl: ptvn