Đặt \(\sqrt{x}=y\\ \) ĐK tồn tại: hiển nhiên\(x\ge0\) và\(\left\{\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}-1\ne0\\\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\ne0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ne4\\x\ne1\\x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}y\ne2\\y\ne1\\y>0\end{matrix}\right.\)bạn chú ý cái đk thứ 3 nhé rất dẽ quên.

\(P=\left(\frac{y^2+3y+2}{\left(y-2\right)\left(y-1\right)}-\frac{y^2+y}{\left(y^2-1\right)}\right):\left(\frac{1}{y+1}+\frac{1}{y-1}\right)\)

\(P=\left(\frac{\left(y^2+3y+2\right)\left(y+1\right)}{\left(y-2\right)\left(y-1\right)\left(y+1\right)}-\frac{\left(y^2+y\right)\left(y-2\right)}{\left(y-2\right)\left(y-1\right)\left(y+1\right)}\right):\left(\frac{y-1+y+1}{\left(y+1\right)\left(y-1\right)}\right)\)

\(P=\left(\frac{\left(y+1\right)\left[\left(y+1\right)\left(y+2\right)-y\left(y-2\right)\right]}{\left(y-2\right)\left(y-1\right)}\right).\left(\frac{\left(y-1\right)\left(y+1\right)}{2y}\right)\)

\(P=\left(\frac{\left(y+1\right)\left(5y+2\right)}{\left(y-2\right)}\right).\left(\frac{\left(y+1\right)}{2y}\right)=\frac{\left(y+1\right)^2\left(5y+2\right)}{2y\left(y-2\right)}\)

sao không gọn đề sai chăng nghi con căn (x)-2 lắm

a) \(P=\frac{\left(\sqrt{x}+1\right)\left(5\sqrt{x}+2\right)}{2\sqrt{x}\left(\sqrt{x}-2\right)}\)

bài này là bài trong sách giáo khoa pk bn

...ghi lại đề...

ĐK: \(x\ge0\)

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

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

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

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

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

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

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

Học tốt! 

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

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

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

có đáp án kĩ hơn không ạ ?

Mấy cái này chỉ có nhân lên rồi rút gọn thôi ah. Nên mình cho bạn đáp án để kiểm tra lại thôi ah

a,Đk: \(\left\{{}\begin{matrix}\sqrt{1+x}\ge0\\\sqrt{1-x}\ge0\\x\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}1+x\ge0\\1-x\ge0\\x\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-1\\x\le1\\x\ne0\end{matrix}\right.\) <=> \(-1\le x\le1,x\ne0\)

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

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

=\(\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}-\frac{\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1-x^2}}{\left|x\right|}-\frac{1-x}{x}\right).\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)( do x>0) (1)

Tại \(0\le x\le1\) => \(\left|x\right|=x\)

Từ (1) <=> A=\(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.\frac{\sqrt{\left(1-x\right)\left(1+x\right)}-\left(1-x\right)}{x}.\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)

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

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

Tại \(-1\le x< 0\)

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

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

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

=-1

Vậy \(\left\{{}\begin{matrix}0\le x\le1\\-1\le x< 0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}A=\frac{1-\sqrt{1-x^2}}{x}\\A=-1\end{matrix}\right.\)

Lê Thị Thục Hiền

Nguyễn Việt Lâm

a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)

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

b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:

\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)

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

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

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

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

Dau "=" xay ra khi x = 0

còn phân b ạ