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Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
hay x=-1
\(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\left(x\ge-5\right)\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\cdot\sqrt{x+5}=6\)
\(\Leftrightarrow3\cdot\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=2^2=4\)
\(\Leftrightarrow x=-1\left(N\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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ĐK \(x\ge-5\)
Ta có: \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow\sqrt{x+5}\left(\sqrt{4}-3+\dfrac{4}{3}\sqrt{9}\right)=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Rightarrow x=\left(\dfrac{6}{3}\right)^2-5=-1\)(TMĐK)
Vậy x=-1
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45=6}\)\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
vậy có tập nghiệm S là -1
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge-5\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9\left(x+5\right)}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\Rightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\) 5 (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)
ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK:5+x\(\ge0\)\(\Leftrightarrow x\ge-5\)
\(\sqrt{4x+20}-3\sqrt{5+x}=6-\dfrac{4}{3}\sqrt{9x+45}\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}=6-\dfrac{4}{3}\sqrt{9\left(x+5\right)}\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}.3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)(tm)
\(\sqrt{4x+20}+3\sqrt{x+5}-\sqrt{9x+45}=6\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}+3\sqrt{x+5}-\sqrt{9\left(x+5\right)}=6\)
\(\Leftrightarrow2\sqrt{x+5}+3\sqrt{x+5}-3\sqrt{x+5}=6\)
\(\Leftrightarrow2\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=3\)
\(\Leftrightarrow x+5=9\)
\(\Leftrightarrow x=4\)