\(ĐK:\left\{{}\begin{matrix}x-2008\ge0\\2008-x\ge0\\x-2007>0\end{matrix}\right.\Leftrightarrow x=2008\)
Vậy PT có nghiệm \(x=2008\)
\(ĐK:\left\{{}\begin{matrix}x-2008\ge0\\2008-x\ge0\\x-2007>0\end{matrix}\right.\Leftrightarrow x=2008\)
Vậy PT có nghiệm \(x=2008\)
giải phương trình vô tỉ sau
\(\left(x-x^2\right)\left(x^2+3x+2007\right)-2005x\sqrt{4-4x}=30\sqrt[4]{x^2+x-1}+2008\)
1`,\(E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)(x>0,x\(\ne\)1)
a,rút gọn E b,Tìm x để E > 0
2,\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right).\left(\sqrt{x}+1\right)\) (x>0,x≠1)
a,rút gọn B b,tìm x để G=2
Rút gọn:
\(A=\left(\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x+\sqrt{x}}\right).\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}-1\)
Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}+3\right)}{x-9}\right)\)\(:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)(với \(x\ge0;x\ne9\))
a) Rút gọn A
b) Tìm x để A<\(-\)1
\(\left(\dfrac{2+\sqrt{X}}{X+\sqrt{X}+1}-\dfrac{\sqrt{X}-2}{x-1}\right)×\dfrac{X\sqrt{X}+x-\sqrt{X}-1}{\sqrt{X}}\left(X>0\:,x\ne1\right)\)
\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right).\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
M = \(\left(\dfrac{\sqrt{x}-2}{x-2\sqrt{x}+1}-\dfrac{\sqrt{x}+2}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)(với x ≥ 0; x≠1)
\(\left(\dfrac{\sqrt{X}-1}{3\sqrt{X}-1}-\dfrac{1}{3\sqrt{X}+1}+\dfrac{8\sqrt{X}}{9X-1}\right):\left(1-\dfrac{3\sqrt{X}-2}{3\sqrt{X}+1}\right)\)