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\(a,ĐK:x\le\dfrac{1}{5}\\ PT\Leftrightarrow1-5x=9\Leftrightarrow x=-\dfrac{8}{5}\\ b,ĐK:x\ge\dfrac{3}{5}\\ PT\Leftrightarrow\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\\sqrt{5x+3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\5x+3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{5}\)
\(c,ĐK:x\ge0\\ PT\Leftrightarrow2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge0\\ PT\Leftrightarrow x-4\sqrt{x}+4-3=0\\ \Leftrightarrow\left(\sqrt{x}-2-\sqrt{3}\right)\left(\sqrt{x}-2+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2+\sqrt{3}\\\sqrt{x}=2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7+4\sqrt{3}\left(tm\right)\\x=7-4\sqrt{3}\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge3\\ PT\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)
Lời giải:
a. ĐKXĐ: $x\leq \frac{1}{5}$
PT $\Leftrightarrow 1-5x=3^2=9$
$\Leftrightarrow 5x=-8\Leftrightarrow x=\frac{-8}{5}$ (tm)
b. ĐKXĐ: $x\geq \frac{3}{5}$
PT $\Leftrightarrow 25x^2-9=4(5x-3)$
$\Leftrightarrow (5x-3)(5x+3)-4(5x-3)=0$
$\Leftrightarrow (5x-3)(5x-1)=0$
$\Leftrightarrow x=\frac{3}{5}$ (tm) hoặc $x=\frac{1}{5}$ (loại)
c. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow x-4\sqrt{x}+3=0$
$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-3)=0$
$\Leftrightarrow \sqrt{x}=1$ hoặc $\sqrt{x}=3$
$\Leftrightarrow x=1$ hoặc $x=9$
d. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-2)^2-5=0$
$\Leftrightarrow (\sqrt{x}-2)^2=5$
$\Leftrightarrow \sqrt{x}-2=\pm \sqrt{5}$
$\Leftrightarrow \sqrt{x}=2+\sqrt{5}$ (chọn) hoặc $\sqrt{x}=2-\sqrt{5}$ (loại do âm)
$\Leftrightarrow x=(2+\sqrt{5})^2=9+4\sqrt{5}$
e.ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow 2\sqrt{9}.\sqrt{x-3}-\frac{1}{5}.\sqrt{25}.\sqrt{x-3}-\frac{1}{7}\sqrt{49}.\sqrt{x-3}=20$
$\Leftrightarrow 6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20$
$\Leftrightarrow 4\sqrt{x-3}=20$
$\Leftrightarrow \sqrt{x-3}=5$
$\Leftrightarrow x-3=25$
$\Leftrightarrow x=28$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(d,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}=2+\sqrt{x+1}\\ \Leftrightarrow x-1=2+x+1+4\sqrt{x+1}\\ \Leftrightarrow4\sqrt{x+1}=-4\Leftrightarrow x\in\varnothing\left(4\sqrt{x+1}\ge0\right)\\ g,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}=2\\ \Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=\dfrac{2-2x}{2}=1-x\\ \Leftrightarrow\left|x-1\right|=1-x\\ \Leftrightarrow\left[{}\begin{matrix}x-1=1-x\left(x\ge1\right)\\x-1=x-1\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x\in R\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{5+2.2.\sqrt{5}+4}{5-4}\)
\(=9+4\sqrt{5}\)
Học tốt!!!!
\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{5+2\sqrt{5}+4+2\sqrt{5}}{\sqrt{5}^2-2^2}\)
\(=\frac{9+4\sqrt{5}}{5-4}=9+4\sqrt{5}\)
@Học tốt@
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\frac{7-2.\sqrt{7}.\sqrt{5}-5}{7-5}\)
\(=\frac{2-2\sqrt{35}}{2}=1-\sqrt{35}\)
Học tốT!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}-\sqrt{9+6\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}=\left(\sqrt{5}-2\right)-\left(3+\sqrt{5}\right)=-5\)
Trả lời:
\(\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{9+6\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(=\sqrt{5}-2-3-\sqrt{5}\)
\(=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)