Bài 1:
a: \(A=\frac15\cdot\sqrt{75}-\sqrt{\left(1-\sqrt3\right)^2}\)
\(=\frac15\cdot5\sqrt3-\left|1-\sqrt3\right|\)
\(=\sqrt3-\left(\sqrt3-1\right)=1\)
b: \(B=\frac{3+\sqrt3}{\sqrt3+1}-\frac{1}{\sqrt3-2}+\sqrt{28-10\sqrt3}\)
\(=\frac{\sqrt3\left(\sqrt3+1\right)}{\sqrt3+1}+\frac{1}{2-\sqrt3}+\sqrt{\left(5-\sqrt3\right)^2}\)
\(=\sqrt3+2+\sqrt3+5-\sqrt3=7+\sqrt3\)
c: \(C=\sin^255^0+\sin^235^0-\tan34^0+\cot56^0-\frac{\cot33^0}{\tan57^0}\)
\(=\left(\sin^255^0+cos^255^0\right)-\tan34^0+\tan34^0-\frac{\tan57^0}{\tan57^0}\)
=1-0-1
=0
Bài 2:
a: ĐKXĐ: 9x-27>=0
=>x>=3
\(\sqrt{9x-27}=1\)
=>\(9x-27=1^2=1\)
=>9x=28
=>\(x=\frac{28}{9}\) (nhận)
b: ĐKXĐ: x>=-1
\(\frac12\cdot\sqrt{4x+4}-2\sqrt{x+1}+5\cdot\sqrt{\frac{4x+4}{25}}=11\)
=>\(\frac12\cdot2\sqrt{x+1}-2\sqrt{x+1}+5\cdot\frac{2\sqrt{x+1}}{5}=11\)
=>\(\sqrt{x+1}-2\sqrt{x+1}+2\sqrt{x+1}=11\)
=>\(\sqrt{x+1}=11\)
=>x+1=11^2=121
=>x=120(nhận)
c: ĐKXĐ: x∈R
\(\sqrt{x^2+2x+1}-1=10x\)
=>\(\sqrt{\left(x+1\right)^2}=10x+1\)
=>|x+1|=10x+1
=>\(\begin{cases}10x+1\ge0\\ \left(10x+1\right)^2=\left(x+1\right)^2\end{cases}\Rightarrow\begin{cases}10x\ge-1\\ \left(10x+1-x-1\right)\left(10x+1+x+1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac{1}{10}\\ 9x\cdot\left(11x+2\right)=0\end{cases}\Rightarrow x\in\left\lbrace0\right\rbrace\)
