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

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

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

\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left\lbrack\frac{-x+\sqrt{x}-1+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x-\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right\rbrack\)

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

Ta có: \(1-\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\right)\cdot\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\)

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

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

Sửa đề: \(A=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\right)\cdot\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}-1\)

Ta có: \(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\)

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

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

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

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

Ta có: \(A=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\right)\cdot\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}-1\)

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

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

Lời giải:

ĐKXĐ: $x>0; x\neq 1$

\(A=\frac{\sqrt{x}-(1-\sqrt{x})}{\sqrt{x}(1-\sqrt{x})}\left[\frac{(2\sqrt{x}-1)(\sqrt{x}+1)}{(1-\sqrt{x})(1+\sqrt{x})}+\frac{\sqrt{x}(2\sqrt{x}-1)(\sqrt{x}+1)}{(1+\sqrt{x})(x-\sqrt{x}+1)}\right]\)

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

Nghe biểu thức cứ sai sai ấy bạn. Có phải giữa 2 ngoặc lớn là dấu chia không?

Ta có: \(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\)

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

Ta có: \(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\)

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

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

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

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

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

Ta có: \(\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right):\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\cdot\sqrt{x}}\right)\)

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

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

c: ĐKXĐ: x>=1/2

Ta có: \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt2\)

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

=>\(\sqrt{2x-1+2\cdot\sqrt{2x-1}\cdot1+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)

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

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

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

=>\(\sqrt{2x-1}-1\le0\)

=>\(\sqrt{2x-1}\le1\)

=>2x-1<=1

=>2x<=2

=>x<=1

=>1/2<=x<=1

d:

ĐKXĐ: x>=-1/4

\(x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4\)

=>\(x+\sqrt{x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14}=4\)

=>\(x+\sqrt{\left(\sqrt{x+\frac14}+\frac12\right)^2}=4\)

=>\(x+\sqrt{x+\frac14}+\frac12=4\)

=>\(x+\frac12+\sqrt{x+\frac14}=4\)

=>\(x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14=4\)

=>\(\left(\sqrt{x+\frac14}+\frac12\right)^2=4\)

=>\(\sqrt{x+\frac14}+\frac12=2\)

=>\(\sqrt{x+\frac14}=2-\frac12=\frac32\)

=>\(x+\frac14=\frac94\)

=>x=2(nhận)

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

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

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

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

1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

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

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

\(=\dfrac{1}{x-1}\)

3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

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

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne9\right)\)

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

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

\(=\dfrac{x+3\sqrt{x}}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{7-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{7-\sqrt{x}}=\dfrac{x}{\sqrt{x}-7}\)

\(B=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\left(x>0,x\ne1\right)\)

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

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

\(=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}}=-\dfrac{1}{\sqrt{x}}\)

a) Ta có: \(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\dfrac{\sqrt{x}}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)

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

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

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

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