1: ĐKXĐ: 3-2x>=0
=>2x<=3
=>\(x< =\dfrac{3}{2}\)
\(\sqrt{3-2x}=5\)
=>3-2x=25
=>2x=3-25=-22
=>\(x=-\dfrac{22}{2}=-11\left(nhận\right)\)
2: ĐKXĐ: 4x-1>=0
=>4x>=1
=>\(x>=\dfrac{1}{4}\)
\(\sqrt{4x-1}+5=7\)
=>\(\sqrt{4x-1}=2\)
=>4x-1=4
=>4x=5
=>\(x=\dfrac{5}{4}\left(nhận\right)\)
3: ĐKXĐ: 3x+5>=0
=>3x>=-5
=>\(x>=-\dfrac{5}{3}\)
\(2\sqrt{3x+5}-3=3\)
=>\(2\sqrt{3x+5}=6\)
=>\(\sqrt{3x+5}=3\)
=>3x+5=9
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
4: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2-28x+49}=2\)
=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot7+7^2}=2\)
=>\(\sqrt{\left(2x-7\right)^2}=2\)
=>|2x-7|=2
=>\(\left[{}\begin{matrix}2x-7=2\\2x-7=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(nhận\right)\\x=\dfrac{5}{2}\left(nhận\right)\end{matrix}\right.\)
6: ĐKXĐ: \(x\in R\)
\(5\sqrt{4x^2-12x+9}-3=7\)
=>\(5\cdot\sqrt{\left(2x\right)^2-2\cdot2x\cdot3+3^2}=10\)
=>\(\sqrt{\left(2x-3\right)^2}=2\)
=>|2x-3|=2
=>\(\left[{}\begin{matrix}2x-3=2\\2x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(nhận\right)\\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
5: \(\sqrt{36+24x+4x^2}-5=1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot6+6^2}=6\)
=>\(\sqrt{\left(2x+6\right)^2}=6\)
=>|2x+6|=6
=>\(\left[{}\begin{matrix}2x+6=6\\2x+6=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
