moi tay

giải giùm mình bài 5 với

mình ko biết

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

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

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

b) \(x=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2\)

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

ý b tự thay vào nha

a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)

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

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

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

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

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

b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)

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

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

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

Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)

c)

Ta có :

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

+)  \(\frac{1}{A}\)nguyên 

\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)

\(\Leftrightarrow x=1\)

Vậy ..............

1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)

Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)

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

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

\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

b) Để \(M=\frac{3}{4}\)

\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)

\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)

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

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

\(\Leftrightarrow x=1\)(tm)

Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)

c) Khi x = 4

\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)

\(\Leftrightarrow M=\frac{2+2}{2+3}\)

\(\Leftrightarrow M=\frac{4}{5}\)

Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)

Cho mik sửa ĐKXĐ: \(x\ne9\)nhé !

Chết : Còn x > 0 nữa nhé

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

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

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

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