Bài 4:
a: ĐKXĐ: x>=0
\(\sqrt{9x}-\sqrt{16x}+\sqrt{81x}=2\)
=>\(3\sqrt{x}-4\sqrt{x}+9\sqrt{x}=2\)
=>\(8\sqrt{x}=2\)
=>\(\sqrt{x}=\frac14\)
=>x=1/16(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}=\sqrt5\)
=>2x-1=5
=>2x=6
=>x=3(nhận)
c: ĐKXĐ: x>=1
\(\frac12\cdot\sqrt{4x-4}-\frac23\cdot\sqrt{9x-9}+24\cdot\sqrt{\frac{x-1}{64}}=6\)
=>\(\frac12\cdot2\sqrt{x-1}-\frac23\cdot3\sqrt{x-1}+24\cdot\frac{\sqrt{x-1}}{8}=6\)
=>\(\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=6\)
=>\(2\sqrt{x-1}=6\)
=>\(\sqrt{x-1}=3\)
=>x-1=9
=>x=10(nhận)
d: \(\sqrt{x^2-4x+4}=3\)
=>\(\sqrt{\left(x-2\right)^2}=3\)
=>|x-2|=3
=>\(\left[\begin{array}{l}x-2=3\\ x-2=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-1\end{array}\right.\)
Bài 3:
a: \(\frac{1}{\sqrt5-2}-\frac{1}{2+\sqrt5}\)
\(=\frac{\sqrt5+2-\left(\sqrt5-2\right)}{\left(\sqrt5+2\right)\left(\sqrt5-2\right)}\)
\(=\frac{\sqrt5+2-\sqrt5+2}{5-4}=\frac41=4\)
b: \(\frac{3}{\sqrt7-2}-\frac{3}{\sqrt7+2}\)
\(=\frac{3\left(\sqrt7+2\right)-3\left(\sqrt7-2\right)}{\left(\sqrt7+2\right)\left(\sqrt7-2\right)}\)
\(=\frac{3\sqrt7+6-3\sqrt7+6}{7-4}=\frac{12}{3}=4\)
c: \(\frac{\sqrt7-7}{\sqrt7-1}=-\frac{\sqrt7\left(\sqrt7-1\right)}{\sqrt7-1}=-\sqrt7\)
d: \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{ab}\)
e: \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{a}}{\sqrt{b}}\)
Bài 1:
a: ĐKXĐ: -7x>=0
=>x<=0
b: ĐKXĐ: 2x+1>=0
=>2x>=-1
=>x>=-1/2
c: ĐKXĐ: 4-3x>=0
=>3x<=4
=>x<=4/3
d: ĐKXĐ: \(\frac{2}{x-1}\ge0\)
=>x-1>0
=>x>1
e: ĐKXĐ: \(x^2+1\ge0\)
mà \(x^2+1\ge1>0\forall x\)
nên x∈R
f: ĐKXĐ: 3-4x>0
=>x<4/3
g: ĐKXĐ: \(\frac{x+1}{2x-3}\ge0\)
=>x>3/2 hoặc x<=-1
