rút gọn bthức sau
\(\left(\frac{2}{2-\sqrt{x}}+\frac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right)\)
D=\(\left(\frac{\sqrt{x}+2}{2-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}-\frac{4x+2\sqrt{x}-4}{x-4}\right)/\left(\frac{2}{2-\sqrt{x}}-\frac{3+\sqrt{x}}{2\sqrt{x}-x}\right)\)
Rút gọn giúp với,...
Cho bt : \(P=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\left(\frac{\sqrt{x}-3}{2\sqrt{x}-x}\right)\)
Rút gọn P
P=\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\left(\frac{\sqrt{x}-3}{2\sqrt{x}-x}\right)=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{4-x}\right).\frac{2\sqrt{x}-x}{\sqrt{x}-3}=\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right].\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{\left(4x+8\sqrt{x}\right).\sqrt{x}.\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\frac{4x}{\sqrt{x}-3}\)
Rút gọn:
\(A=\left(\frac{4x\sqrt{x}+3x+9}{x+5\sqrt{x}+6}-\frac{3-\sqrt{x}}{2+\sqrt{x}}\right)\div\left(\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{3+4\sqrt{x}}{x+5\sqrt{x}+6}\right)\)
\(B=\left(x-\sqrt{x}-2\right)\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{4-\sqrt{x}}{x-2\sqrt{x}}\right)\)
Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
rút gọn biểu thức sau:
H= \(\left(\frac{\sqrt{x}+4}{x-2\sqrt{x}}+\frac{3}{\sqrt{x}-2}\right)\div\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\)
Rút gọn:
\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
= \(\left[\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
= \(\left[\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-3\right)^2}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)
= \(\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\frac{\sqrt{x}-3}{2-\sqrt{x}}\)
= \(\left[\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{nt}\right]:nt\)
\(=\left[\frac{8\sqrt{x}+4x}{nt}\right]:nt\)
\(=\left[\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:nt\)
\(=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(=\frac{4\sqrt{x}}{\sqrt{x}-3}\)
Rút gọn B=\(\left(\frac{1}{\sqrt{x}-2}-\frac{2}{\sqrt{x}+2}+\frac{x}{x\sqrt{x}-4\sqrt{x}}\right):\left(\frac{6-x}{\sqrt{x}+2}+2+\sqrt{x}\right)\)
\(B=\left(\frac{1}{\sqrt{x}-2}-\frac{2}{\sqrt{x}+2}+\frac{x}{x\sqrt{x}-4\sqrt{x}}\right):\left(\frac{6-x}{\sqrt{x}+2}+2+\sqrt{x}\right)\)
\(B=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\right):\left(\frac{6-x+2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(B=\left(\frac{\sqrt{x}+2-2\sqrt{x}+4+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{6-x+2\sqrt{x}+4+x+2\sqrt{x}}{\sqrt{x}+2}\right)\)
\(B=\frac{6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{10+4\sqrt{x}}\)
\(B=\frac{6}{\sqrt{x}-2}\cdot\frac{1}{2\left(5+2\sqrt{x}\right)}\)
B = \(\frac{3}{\left(\sqrt{x}-2\right)\left(5+2\sqrt{x}\right)}\)
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
Rút gọn biểu thức:
\(Q=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{1}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x-4}\right)\)
đkxđ: \(x\ge0;x\ne4\)
\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\div\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)
\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)
\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)
đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)