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Lời giải:
a) ĐK: \(x>0; x\neq 25; x\neq 36\)
PT \(\Rightarrow (\sqrt{x}-2)(\sqrt{x}-6)=(\sqrt{x}-5)(\sqrt{x}-4)\)
\(\Leftrightarrow x-8\sqrt{x}+12=x-9\sqrt{x}+20\)
\(\Leftrightarrow \sqrt{x}=8\Rightarrow x=64\) (thỏa mãn)
Vậy.......
b)
ĐK: \(x\geq \frac{-1}{2}\)
PT \(\Leftrightarrow \sqrt{9(2x+1)}-\sqrt{4(2x+1)}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow 3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow \frac{4}{3}\sqrt{2x+1}=4\Leftrightarrow \sqrt{2x+1}=3\)
\(\Rightarrow x=\frac{3^2-1}{2}=4\) (thỏa mãn)
c)
ĐK: \(x\geq 2\)
PT \(\Leftrightarrow \sqrt{4(x-2)}-\frac{1}{2}\sqrt{x-2}+\sqrt{9(x-2)}=9\)
\(\Leftrightarrow 2\sqrt{x-2}-\frac{1}{2}\sqrt{x-2}+3\sqrt{x-2}=9\)
\(\Leftrightarrow \frac{9}{2}\sqrt{x-2}=9\Leftrightarrow \sqrt{x-2}=2\Rightarrow x=2^2+2=6\) (thỏa mãn)
1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy x=2 hoặc x=-1
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
ĐKXĐ:
a/ \(x-2020>0\Rightarrow x>2020\)
b/ \(x\ne0\)
c/ \(3x+5< 0\Rightarrow x< -\frac{5}{3}\)
d/ \(\frac{x-3}{1-x}\ge0\Rightarrow1< x\le3\)
Bài 2: ĐKXĐ tự tìm
a/ \(2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\Rightarrow\sqrt{2x}=\frac{28}{13}\)
\(\Rightarrow x=\frac{392}{169}\)
b/ \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\Rightarrow x=9\)
c/ \(3\sqrt{2x+1}>15\Rightarrow\sqrt{2x+1}>5\)
\(\Rightarrow2x+1>25\Rightarrow x>12\)
d/ \(\sqrt{x}+1>12\Rightarrow\sqrt{x}>11\Rightarrow x>121\)