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3 tháng 8 2017

Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:

\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)

=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)

=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)

3 tháng 8 2017

bùn ngủ , mai lm câu b cho nha

12 tháng 8 2019

a) đk : \(x\ge0\) ; \(x\ne1\)

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

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

b) đk : \(x\ne0;x\ne1\)

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

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

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

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

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

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

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

1 tháng 8 2018

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

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

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

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

8 tháng 11 2020

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

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

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

Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)

9 tháng 11 2020

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

Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0

AH
Akai Haruma
Giáo viên
2 tháng 6 2020

Lời giải:

a)

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

b)

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

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

13 tháng 9 2019

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

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

Ta co:

\(\sqrt{x}-1+\frac{2}{\sqrt{x}}=\frac{x-\sqrt{x}+2}{\sqrt{x}}=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}{\sqrt{x}}>0\)

\(\Rightarrow\sqrt{x}-1>-\frac{2}{\sqrt{x}}\)

7 tháng 7 2018

mk làm luôn

a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)

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

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

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

=

6 tháng 7 2018

a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)

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

\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)

\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)

\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)

\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)

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

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

\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)

\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)

\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)

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

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

7 tháng 7 2018

bạn huy hoàng làm sai rồi