a: ĐKXĐ: \(\frac{2x-3}{x-1}\ge0\)
=>x>=3/2 hoặc x<1
Ta có: \(\sqrt{\frac{2x-3}{x-1}}=2\)
=>\(\frac{2x-3}{x-1}=4\)
=>4(x-1)=2x-3
=>4x-4=2x-3
=>4x-2x=-3+4
=>2x=1
=>\(x=\frac12\) (nhận)
b: ĐKXĐ: \(\frac{x-3}{2x+1}\ge0\)
=>x>=3 hoặc x<-1/2
\(\sqrt{\frac{x-3}{2x+1}}=5\)
=>\(\frac{x-3}{2x+1}=25\)
=>25(2x+1)=x-3
=>50x+25=x-3
=>49x=-28
=>\(x=-\frac{28}{49}=-\frac47\) (nhận)
c: ĐKXĐ: \(\begin{cases}4x^2-9\ge0\\ 2x+3\ge0\end{cases}\Rightarrow\begin{cases}4x^2\ge9\\ x\ge-\frac32\end{cases}\)
=>x>=-3/2 và (x>=3/2 hoặc x<=-3/2)
=>(x=-3/2 hoặc x>=3/2)
\(\sqrt{4x^2-9}=2\cdot\sqrt{2x+3}\)
=>\(\sqrt{\left(2x-3\right)\left(2x+3\right)}-2\cdot\sqrt{2x+3}=0\)
=>\(\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
TH1: \(\sqrt{2x+3}=0\)
=>2x+3=0
=>2x=-3
=>\(x=-\frac32\) (nhận)
TH2: \(\sqrt{2x-3}-2=0\)
=>\(\sqrt{2x-3}=2\)
=>2x-3=4
=>2x=7
=>x=7/2(nhận)
d:ĐKXĐ: x>=2
\(\sqrt{x-2}-\sqrt{x^2-4}=0\)
=>\(\sqrt{x-2}\left(1-\sqrt{x+2}\right)=0\)
TH1: \(\sqrt{x-2}=0\)
=>x-2=0
=>x=2(nhận)
Th2: \(1-\sqrt{x+2}=0\)
=>\(\sqrt{x+2}=1\)
=>x+2=1
=>x=-1(loại)
e: \(\sqrt{x^2+8x+16}=5\)
=>\(\sqrt{\left(x+4\right)^2}=5\)
=>|x+4|=5
=>\(\left[\begin{array}{l}x+4=5\\ x+4=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5-4=1\\ x=-5-4=-9\end{array}\right.\)
f: \(\sqrt{x^2-10x+25}=7\)
=>\(\sqrt{\left(x-5\right)^2}=7\)
=>|x-5|=7
=>\(\left[\begin{array}{l}x-5=7\\ x-5=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7+5=12\\ x=-7+5=-2\end{array}\right.\)
g: \(\sqrt{x^2+6x+9}=3x-6\)
=>\(\sqrt{\left(x+3\right)^2}=3x-6\)
=>|x+3|=3x-6
=>\(\begin{cases}3x-6\ge0\\ \left(3x-6\right)^2=\left(x+3\right)^2\end{cases}\Rightarrow\begin{cases}3x\ge6\\ \left(3x-6-x-3\right)\left(3x-6+x+3\right)=0\end{cases}\)
=>\(\begin{cases}x\ge2\\ \left(2x-9\right)\left(4x-3\right)=0\end{cases}\Rightarrow x=\frac92\)
h: \(\sqrt{x^2-4x+4}-2x+5=0\)
=>\(\sqrt{\left(x-2\right)^2}=2x-5\)
=>|x-2|=2x-5
=>\(\begin{cases}2x-5\ge0\\ \left(2x-5\right)^2=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac52\\ \left(2x-5-x+2\right)\left(2x-5+x-2\right)=0\end{cases}\)
=>\(\begin{cases}x\ge\frac52\\ \left(x-3\right)\left(3x-7\right)=0\end{cases}\Rightarrow x=3\)
i: ĐKXĐ: x-3>=0 và 2x+1>0
=>x>=3
\(\frac{\sqrt{x-3}}{\sqrt{2x+1}}=2\)
=>\(\sqrt{\frac{x-3}{2x+1}}=2\)
=>\(\frac{x-3}{2x+1}=4\)
=>4(2x+1)=x-3
=>8x+4=x-3
=>7x=-7
=>x=-1(loại)
j: ĐKXĐ: x>-5/7
\(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
=>9x-7=7x+5
=>2x=12
=>x=6(nhận)
k: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\frac13\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\frac13\cdot3\cdot\sqrt{x-5}=4\)
=>\(2\cdot\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
l: ĐKXĐ: x>=3
\(2\cdot\sqrt{9x-27}-\frac15\cdot\sqrt{25x-75}-\frac17\cdot\sqrt{49x-147}=20\)
=>\(2\cdot3\cdot\sqrt{x-3}-\frac15\cdot5\sqrt{x-3}-\frac17\cdot7\cdot\sqrt{x-3}=20\)
=>\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
=>\(4\cdot\sqrt{x-3}=20\)
=>\(\sqrt{x-3}=5\)
=>x-3=25
=>x=28(nhận)
