a: \(\frac{\sqrt6+\sqrt{14}}{2\sqrt3+\sqrt{28}}\)
\(=\frac{\sqrt2\left(\sqrt3+\sqrt7\right)}{2\sqrt3+2\sqrt7}=\frac{\sqrt2\left(\sqrt3+\sqrt7\right)}{2\left(\sqrt3+\sqrt7\right)}=\frac{\sqrt2}{2}\)
b: \(\frac{\sqrt2+\sqrt3+\sqrt6+\sqrt8+\sqrt{16}}{\sqrt2+\sqrt3+\sqrt4}\)
\(=\frac{\sqrt2+\sqrt3+2+2+\sqrt6+\sqrt8}{\sqrt2+\sqrt3+2}=\frac{\left(\sqrt2+\sqrt3+2\right)+\sqrt2\left(\sqrt2+\sqrt3+2\right)}{\sqrt2+\sqrt3+2}\)
\(=1+\sqrt2\)
c:Đặt A= \(\sqrt{\frac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\frac{x^2-1}{x-3}\)
\(=\left|\frac{x-2}{x-3}\right|+\frac{x^2-1}{x-3}\)
\(=\frac{\left|x-2\right|}{3-x}+\frac{x^2-1}{x-3}\) (Vì x<3)
TH1: x<=2
=>\(A=\frac{2-x}{3-x}+\frac{x^2-1}{x-3}\)
\(=\frac{x-2+x^2-1}{x-3}=\frac{x^2+x-3}{x-3}\)
TH2: 2<=x<3)
=>\(A=\frac{x-2}{3-x}+\frac{x^2-1}{x-3}=\frac{-x+2+x^2-1}{x-3}=\frac{x^2-x+1}{x-3}\)
d: Đặt B=\(4x-\sqrt8+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt2+\sqrt{\frac{x^2\left(x+2\right)}{x+2}}\)
\(=4x-2\sqrt2+\left|x\right|\)
TH1: x>=0
=>\(\left|x\right|=x\)
\(B=4x-2\sqrt2+\left|x\right|\)
\(=4x-2\sqrt2+x=5x-2\sqrt2\)
TH2: -2<x<0
=>|x|=-x
=>\(B=4x-2\sqrt2-x=3x-2\sqrt2\)
giúp e với ạ e cảm ơnn
