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`a)\sqrt{16x+48}+\sqrt{x+3}=15` `ĐK: x >= -3`
`<=>4\sqrt{x+3}+\sqrt{x+3}=15`
`<=>5\sqrt{x+3}=15`
`<=>\sqrt{x+3}=3`
`<=>x+3=9<=>x=6` (t/m).
`b)\sqrt{x^2-4}-3\sqrt{x-2}=0` `ĐK: x >= 2`
`<=>\sqrt{x-2}(\sqrt{x+2}-3)=0`
`<=>[(\sqrt{x-2}=0),(\sqrt{x+2}=3):}`
`<=>[(x-2=0),(x+2=9):}<=>[(x=2(t//m)),(x=7(t//m)):}`
![](https://rs.olm.vn/images/avt/0.png?1311)
a, bạn viết rõ đề được ko ?
b, \(x-5\sqrt{x}-1+5=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-5\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\Leftrightarrow x=1;x=2\)
Vậy tập nghiệm của phương trình là S = { 1 ; 2 }
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>x-2=0 hoặc x+2=1/9
=>x=-17/9(loại) hoặc x=2
b: \(\Leftrightarrow\sqrt{x^2-1}\left(1-\sqrt{x^2-1}\right)=0\)
=>x^2-1=0 hoặc x^2-1=1
=>x^2=1 hoặc x^2=2
=>\(x\in\left\{1;-1;\sqrt{2};-\sqrt{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. ĐKXĐ: $x\in\mathbb{R}$
$\sqrt{4(x+2)^2}=8$
$\Leftrightarrow 2\sqrt{(x+2)^2}=8$
$\Leftrightarrow \sqrt{(x+2)^2}=4$
$\Leftrightarrow |x+2|=4\Rightarrow x+2=\pm 4$
$\Rightarrow x=2$ hoặc $x=-6$ (đều thỏa mãn)
b. ĐKXĐ: $x\in\mathbb{R}$
PT \(\sqrt{(x-3)^2}=3-x\Leftrightarrow |x-3|=3-x\Leftrightarrow 3-x\geq 0\Leftrightarrow x\leq 3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x-4+3y+3=-2 và 3x-6+2y+2=-3
=>2x+3y=-2-3+4=-1 và 3x+2y=-3+6-2=1
=>x=1;y=-1
b: =>1/2x=4/3 và 5x-8y=3
=>x=4/3:1/2=4/3*2=8/3 và 8y=5x-3=5*8/3-3=40/3-3=31/3
=>y=31/24; x=8/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\sqrt{2x+5}=\sqrt{1-x}\)
<=> 2x + 5 = 1 - x
<=> 2x + x = 1 - 5
<=> 3x = -4
<=> x = \(\dfrac{-4}{3}\)
Vậy ...............
b. \(\sqrt{x^2-x}=\sqrt{3-x}\)
<=> x2 - x = 3 - x
<=> x2 - x + x = 3
<=> x2 = 3
<=> x = \(\sqrt{3}\)
Vậy ..................
c. \(\sqrt{2x^2-3}=\sqrt{4x-3}\)
<=> 2x2 - 3 = 4x - 3
<=> 2x2 - 4x = -3 + 3
<=> 2x2 - 4x = 0
<=> x(x - 4) = 0
\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy .................
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{\left(x-3\right)^2}=2\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\left(đk:x\ge-2\right)\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\Leftrightarrow x+2=9\Leftrightarrow x=7\)
a: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
a,ĐK: x≥4
Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}=4\)
\(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)
b, ĐK: x≥2
Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=5$
$\Leftrightarrow |x-2|=5$
$\Leftrightarrow x-2=5$ hoặc $x-2=-5$
$\Leftrightarrow x=7$ hoặc $x=-3$ (đều tm)
b. ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow \sqrt{16}.\sqrt{x+1}-3\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}=16-\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Leftrightarrow x+1=16$
$\Leftrightarrow x=15$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left\{{}\begin{matrix}x+y\ne0\\x-y\ne0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}2u+v=3\\u-3v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6u+3v=9\\u-3v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7u=10\\u-3v=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{10}{7}\\v=\dfrac{1}{7}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=\dfrac{10}{7}\\\dfrac{1}{x-y}=\dfrac{1}{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{7}{10}\\x-y=7\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{77}{20}\\y=-\dfrac{63}{20}\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{2}{x+y}+\dfrac{1}{x-y}=3\\\dfrac{1}{x+y}-\dfrac{3}{x-y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}+\dfrac{1}{x-y}=3\\\dfrac{2}{x+y}-\dfrac{6}{x-y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{x-y}=1\\\dfrac{1}{x+y}-\dfrac{3}{x-y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=7\\\dfrac{1}{x+y}=1+\dfrac{3}{x-y}=\dfrac{10}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=7\\x+y=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{77}{10}\\x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{77}{20}\\y=\dfrac{-63}{20}\end{matrix}\right.\)