Question

[Tran Thu Hong]

Rút gọn bt:

a, M=\(\sqrt{2}+1-\sqrt{3-2\sqrt{2}}\)

b, N=\(\sqrt{1+2\sqrt{3+2\sqrt{2}}}\)

c,P= \(\frac{x-1}{\sqrt{x}+1}+\frac{2x}{\sqrt{x}}\)

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

Answer 1

[Tuyển Trần Thị]

M= \(\sqrt{2}+1-\) \(\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\)

N=\(\sqrt{1+2\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{1+2\left(\sqrt{2}+1\right)}=\) \(\sqrt{1+2\sqrt{2}+2}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

P= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}}\) (dk \(x>0\))

=\(\sqrt{x}+1+2\sqrt{x}=3\sqrt{x}+1\)

Q= \(\sqrt{\left(\sqrt{x}+1\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\) (dk \(x\ge0\) )

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

th1 \(\sqrt{x}\ge1\Leftrightarrow x\ge1\) Q=\(\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)

th2 \(0\le x< 1\) Q=\(\sqrt{x}+1+1-\sqrt{x}=2\)

Answer 2

[Hàn Tiểu Lam]

a)  \(M=\sqrt{2}+1-\sqrt{1,5.2-2.\sqrt{2}}\)

\(=\sqrt{2}+1-\sqrt{2.\left(1,5-\sqrt{2}\right)}\)\(=\sqrt{2}+1-\sqrt{2}.\sqrt{1,5-\sqrt{2}}\)

\(=\sqrt{2}.\left(1+1,5-\sqrt{2}\right)+1=\sqrt{2}.\left(2,5-\sqrt{2}\right)+1\)

\(=\sqrt{2}.2,5-2+1=\sqrt{2}.2,5-1\)

P/s: Theo em thì em nghĩ là đúng '-' Khoảng 90% :)