Giải phương trình sqrt(x - 2) + 2sqrt(9x - 18) - sqrt(4x - 8) = 10 giúp e ạ
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![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow 2\sqrt{x+2}+3\sqrt{4}.\sqrt{x+2}-\sqrt{9}.\sqrt{x+2}=10$
$\Leftrightarrow 2\sqrt{x+2}+6\sqrt{x+2}-3\sqrt{x+2}=10$
$\Leftrightarrow 5\sqrt{x+2}=10$
$\Leftrightarrow \sqrt{x+2}=2$
$\Leftrightarrow x+2=4$
$\Leftrightarrow x=2$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: ĐKXĐ: x>=2
\(2\sqrt{9x-18}-\sqrt{x-2}+\dfrac{1}{2}\cdot\sqrt{4x-8}=18\)
=>\(2\cdot3\cdot\sqrt{x-2}-\sqrt{x-2}+\dfrac{1}{2}\cdot2\sqrt{x-2}=18\)
=>\(6\sqrt{x-2}=18\)
=>\(\sqrt{x-2}=3\)
=>x-2=9
=>x=11(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)
\(\Leftrightarrow\sqrt{x+3}=3\)
\(\Leftrightarrow x+3=9\)
hay x=6
b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge2\)
Ta có: \(\sqrt{4x-8}-\sqrt{9x-18}+2\sqrt{x-2}=1\)
\(\Leftrightarrow2\sqrt{x-2}-3\sqrt{x-2}+2\sqrt{x-2}=1\)
\(\Leftrightarrow\sqrt{x-2}=1\)
\(\Leftrightarrow x-2=1\)
hay x=3(nhận)
Vậy: S={3}
![](https://rs.olm.vn/images/avt/0.png?1311)
a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)
\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)
\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\sqrt{x-2}-\sqrt{4x-8}+4\sqrt{\dfrac{9x-18}{4}}=14\left(x\ge0;x\ne2\right)\\ \Leftrightarrow3\sqrt{x-2}-\sqrt{4\left(x-2\right)}+4\cdot\dfrac{1}{2}\sqrt{9\left(x-2\right)}=14\\ \Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+6\sqrt{x-2}=14\\ \Leftrightarrow7\sqrt{x-2}=14\\ \Leftrightarrow\sqrt{x-2}=2\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\)
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)
\(a,ĐK:x\ge2\\ PT\Leftrightarrow4\sqrt{x-2}-2\sqrt{x-2}-7\sqrt{x-2}=-10\\ \Leftrightarrow-5\sqrt{x-2}=-10\\ \Leftrightarrow\sqrt{x-2}=2\Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\\ b,ĐK:x\ge1\\ PT\Leftrightarrow x-3=\sqrt{x-1}\\ \Leftrightarrow x^2-6x+9=x-1\\ \Leftrightarrow x^2-7x+10=0\\ \Leftrightarrow\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\left(tm\right)\)
ĐKXĐ: x ≥ 2
Phương trình đã cho tương đương:
√(x - 2) + 6√(x - 2) - 2√(x - 2) = 10
⇔ 5√(x - 2) = 10
⇔ √(x - 2) = 2
⇔ x - 2 = 4
⇔ x = 6 (nhận)
Vậy S = {6}