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

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

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

cvfbhm,

Xin lỗi em ko biết làm , em vẫn chưa lên lớp 9

Bạn ơi, bạn ghi lại đề đi bạn. Khó hiểu quá!

Đề là \(x+y-\sqrt{xy}=3\) với \(\sqrt{x+1}+\sqrt{y-1}=4\) pk bạn?

\(\sqrt{x^3-6x^2+12x-8}\)

\(=\sqrt{\left(x-2\right)^3}\)

\(=\left|x-2\right|\cdot\sqrt{x-2}\)

a: \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b: Thay x=9 vào P, ta được:

\(P=\frac{\sqrt9-1}{\sqrt9+1}=\frac{3-1}{3+1}=\frac24=\frac12\)

\(\sqrt{4}+\sqrt{15}-\sqrt{4}-\sqrt{15}-\sqrt{2}-\sqrt{3}\)

\(=-\sqrt{3}-\sqrt{2}\)

1: \(T=\left(1-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{x-1}\)

\(=\frac{\sqrt{x}+1-1}{\sqrt{x}+1}:\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}+1\right)=\sqrt{x}\)

2: \(x=17-12\sqrt2=\left(3-2\sqrt2\right)^2\)

=>\(T=\sqrt{\left(3-2\sqrt2\right)^2}=3-2\sqrt2\)

=>\(T+2\sqrt2=3\)

Bằng không!

\(B=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

\(=\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

\(=|1+\sqrt{3}|-|1-\sqrt{3}|\)

\(=1+\sqrt{3}-\left(\sqrt{3}-1\right)\)

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