\(P=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

ĐKXĐ:\(x\ge0;x\ne9\)

\(=\left(\frac{x+3}{x-9}+\frac{1\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\left(\frac{x+3+\sqrt{x}-3}{x-9}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{x+\sqrt{x}}{x-9}.\frac{\sqrt{x-3}}{\sqrt{x}}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b)

\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=\sqrt{5^2+2.5\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{4^2+2.4\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

\(=1\)

Thay x=1 vào P ta có:

\(P=\frac{\sqrt{1}+1}{\sqrt{1}-3}\)

\(=\frac{2}{-2}=-1\)

Có bị sai đề không vậy bạn ? Mình nghĩ nó là \(\sqrt{x}+3\) với \(\sqrt{x}-3\)chứ không phải là \(\sqrt{x+3}\) với \(\sqrt{x-3}\)?

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

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

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

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

a) ĐK: x > 1

 \(P=\left(\frac{\sqrt{x-1}}{3+\sqrt{x-1}}+\frac{x+8}{9-\left(x-1\right)}\right):\left(\frac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\frac{1}{\sqrt{x-1}}\right)\)

\(P=\frac{\sqrt{x-1}\left(3-\sqrt{x-1}\right)+x+8}{9-\left(x-1\right)}:\frac{3\sqrt{x-1}+1-\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)

\(P=\frac{3\sqrt{x-1}-x+1+x+8}{10-x}:\frac{2\sqrt{x-1}+4}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)

\(P=\frac{3\left(\sqrt{x-1}+3\right)}{10-x}.\frac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}{2\sqrt{x-1}+4}\)

\(P=\frac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)

b) \(x=\sqrt[4]{\frac{17+12\sqrt{2}}{1}}-\sqrt[4]{\frac{17-12\sqrt{2}}{1}}=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)

Vậy \(P=\frac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\frac{1}{2}\)

cô Hoàng Thị Thu Huyền làm rõ cho em ý b đc ko ạ chỗ biến đổi x 

tất cả phải nghe theo lệnh của cô chủ nhiệm tập đoàn của cô Hoàng Thị Thu Huyền&Trần Thị Loan nhưng hôm nay không cô Trần Thị Loan nên tôi thay mặt cho vì hai cô này có chức Quản Lý

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

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

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

\(=1-a+a=1\)