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24 tháng 9 2019

\(A=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right):\frac{x+2\sqrt{x}}{\sqrt{x}}.\)

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

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

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

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

\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}}{x+2\sqrt{x}}\)

\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x+2}\right)}\)

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

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

22 tháng 12 2017

\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)  \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)

\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)

\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)

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

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

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

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

b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)

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

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

\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)

\(\Leftrightarrow x-2\sqrt{x}-2=0\)

\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)

vậy......

31 tháng 7 2017

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

\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{2x}{\sqrt{x}-2}\)

7 tháng 4 2019

a=căn (x)

A=[(4-a^2)(2-a)+2a(4-a^2)-4a^2(2-a)]/[(4-a^2)(2-a)2a]

A=(8-10a^2+4a+3a^3)/a(16-4a^2-8a+2a^3)

A=(a-2)^2(3a+2)/a(a+2)(a-2)^2*2

A=(3a+2)/a(a+2)*2

B=2+căn(3)

A=B suy ra

(3a+2)/a(a+2)*2=2+căn 3

<=>bấm máy tính ra nghiệm a=0.1539181357

=>x=a^2 =0.02341454985

tl đúng

4 tháng 8 2019

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

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

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

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

\(b,A=0\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=0\Leftrightarrow\sqrt{x}-2=0\)

Mà \(\sqrt{x}+2\ne0\)\(\Rightarrow\)không có giá trị nào  của x thỏa mãn \(A=0\)

29 tháng 6 2016

đề có sai không zậy?