Đáp án:
D= $\frac{\sqrt[]{x^2}+\sqrt[]{4-4x+x^2}}{x-1}$
D=$\frac{lxl+\sqrt[]{(x-2)^2}}{x-1}$
D=$\frac{lxl+lx-2l}{x-1}$
TH1 x≥0
x+2≥0 => x≥0
=>D=$\frac{x+x-2l}{x-1}$ =2
TH2 x≥0
x+2≤0( vô lí )
TH3 x≤0
x+2≤0
=>x≤-2
=>D = $\frac{-x-x+2}{x-1}$ =-2
TH4
x≤0
x+2≥0
=> -2≤x≤0
=> D=$\frac{-x+x-2}{x-1}$ =$\frac{-2}{x-1}$
E=$\sqrt[]{a+4\sqrt[]{a-2}+2}$ +$\sqrt[]{a-4\sqrt[]{a-2}+2}$
E=$\sqrt[]{a-2+4\sqrt[]{a-2}+4}$ +$\sqrt[]{a-2-4\sqrt[]{a-2}+4}$
E=$\sqrt[]{(\sqrt[]{a-2}+2)^2}$ +$\sqrt[]{(\sqrt[]{a-2}-2)^2}$
E=l$\sqrt[]{a-2}$ +2 l+l$\sqrt[]{a-2}$ -2l
TH1 $\sqrt[]{a-2}$ -2≥0
=> a-2≥4
=.a≥6
E= $\sqrt[]{a-2}$ +2 +$\sqrt[]{a-2}$ -2=2$\sqrt[]{a-2}$
TH2 $\sqrt[]{a-2}$ -2≤0
=> a-2≤4
=. a≤6
=> E=$\sqrt[]{a-2}$ +2 -$\sqrt[]{a-2}$ +2=4