Bài 3:
a: \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\cdot\frac{x-1}{2\left(x-2\sqrt{x}+1\right)}\)
\(=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right)\cdot\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: \(B=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Bài 2:
a: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{x+3}=7\)
=>\(x+3=7^2=49\)
=>x=49-3
=>x=46(nhận)
b; ĐKXĐ: x>=0
\(\sqrt{9x}+\sqrt{25x}-\sqrt{x}=28\)
=>\(3\sqrt{x}+5\sqrt{x}-\sqrt{x}=28\)
=>\(7\sqrt{x}=28\)
=>\(\sqrt{x}=4\)
=>x=16(nhận)
c: ĐKXĐ: x>=1
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{x-1}=5\)
=>\(4\sqrt{x-1}-3\sqrt{x-1}+\sqrt{x-1}=5\)
=>\(2\sqrt{x-1}=5\)
=>\(\sqrt{x-1}=\frac52\)
=>\(x-1=\frac{25}{4}\)
=>\(x=\frac{25}{4}+1=\frac{29}{4}\) (nhận)
d: ĐKXĐ: x∈R
\(\sqrt{\left(x-1\right)^2}=2\)
=>|x-1|=2
=>\(\left[\begin{array}{l}x-1=2\\ x-1=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-1\end{array}\right.\)
e: \(\sqrt{x^2-2x+1}-7=0\)
=>\(\sqrt{\left(x-1\right)^2}=7\)
=>|x-1|=7
=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)
