Bài 3:
a: \(\sqrt{36x^2-60x+25}=4\)
=>\(\sqrt{\left(6x-5\right)^2}=4\)
=>|6x-5|=4
=>\(\left[\begin{array}{l}6x-5=4\\ 6x-5=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}6x=9\\ 6x=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32\\ x=\frac16\end{array}\right.\)
b:
ĐKXĐ: x>=5
\(\sqrt{4x-20}+3\cdot\sqrt{\frac{x-5}{9}}-\frac14\cdot\sqrt{16x-80}=6\)
=>\(2\cdot\sqrt{x-5}+3\cdot\frac{\sqrt{x-5}}{3}-\frac14\cdot4\cdot\sqrt{x-5}=6\)
=>\(2\cdot\sqrt{x-5}=6\)
=>\(\sqrt{x-5}=3\)
=>x-5=9
=>x=9+5=14(nhận)
c: ĐKXĐ: 5-2x>=0
=>2x<=5
=>x<=5/2
\(\sqrt{5-2x}=3-x\)
=>\(\begin{cases}3-x\ge0\\ \left(3-x\right)^2=5-2x\end{cases}\Rightarrow\begin{cases}x\le3\\ x^2-6x+9+2x-5=0\end{cases}\)
=>\(\begin{cases}x\le3\\ \left(x-2\right)^2=0\end{cases}\)
=>x=2
Bài 2:
a: \(\frac13\cdot\sqrt{153}=\sqrt{\frac{153}{9}}=\sqrt{17}\)
\(3\sqrt2=\sqrt{3^2\cdot2}=\sqrt{18}\)
mà 17<18
nên \(\frac13\cdot\sqrt{153}<3\sqrt2\)
b:ĐKXĐ: \(5-2x\ge0\)
=>2x<=5
=>x<=5/2
c: ĐKXĐ: \(x\ge2\sqrt2\)
\(\sqrt{x-2\sqrt2}=\sqrt2-1\)
=>\(x-2\sqrt2=\left(\sqrt2-1\right)^2=3-2\sqrt2\)
=>x=3(nhận)
