Question

Rút gọn.

a) \(\frac{x-y}{\sqrt{x}-\sqrt{y}}\) với x ≥0 , y ≥ 0, x≠ y

b)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}\) với x ≥0 , x≠ 1

c)\(\sqrt{x+1\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\) với x ≥ 1

d)\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)

Answer 1

c,C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\left(x\ge1\right)\)

=\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

=\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\) (1)

TH1: \(\sqrt{x-1}< 1\) hay \(1\le x< 2\)

Từ (1)=>C= \(\sqrt{x-1}+1+1-\sqrt{x-1}\)=2

TH2: \(\sqrt{x-1}\ge1\) hay \(x\ge2\)

Từ (1) =>C=\(\sqrt{x-1}+1+\sqrt{x-1}-1\)=\(2\sqrt{x-1}\)

d, D=\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{13+30\sqrt{2}+\sqrt{8+2\sqrt{8}+1}}=\sqrt{13+30\sqrt{2}+\sqrt{\left(\sqrt{8}+1\right)^2}}\)

=\(\sqrt{13+30\sqrt{2}+\sqrt{8}+1}=\sqrt{14+30\sqrt{2}+2\sqrt{2}}\)

=\(\sqrt{14+32\sqrt{2}}\)

Answer 2

a)\(\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)

b)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)