\(\sqrt{2x^2+x-1}-x\sqrt{x+1}-2x-1=x\sqrt{2x-1}\)
Điều kiện: \(x\ge\dfrac{1}{2}\)
\(\Leftrightarrow-2\sqrt{2x^2+x-1}+2x\sqrt{x+1}+4x+2+2x\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-\sqrt{x+1}\right)^2+2x\sqrt{x+1}+x+2+2x\sqrt{2x-1}=0\)
Ta có \(VT>0\) nên vô nghiệm
Rút gọn : A=\(\dfrac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}.\left(\sqrt{2x-1}\right)\)
Giải phương trình:
a) \(2\left(1-x\right)\sqrt{2x^2+2x-1}=x^2-x+1\)
b) \(\sqrt{x-1}+x-1=\sqrt{2\left(x-3\right)^2+2x-2}\)
c) \(\sqrt{x^2+2x}+\sqrt{2x-1}=\sqrt{3x^2+4x+1}\)
d) \(\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) (\(x\ge0;x\ne4\))
B = \(\frac{x\sqrt{x}-1}{x-1}\) (\(x\ge0;x\ne1\))
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) ( \(x>0;x\ne1\))
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (\(x\ge2\))
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2}-2x}\)
1) \(2\sqrt{x+3}=x-1+4\sqrt{2x-1}\)
2) \(\sqrt[4]{x-1}+\sqrt[4]{5-x}=2\)
3) \(\sqrt[3]{1-2x}+\sqrt{x+3}=1\)
4) \(\dfrac{\sqrt{x}}{1+\sqrt{1-x}}=x^2-2x+2\)
Rút gọn: \(F=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2x-1}\)
Rút gọn : \(Q=\frac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}}\)với \(x\ge2\)
Giải phương trình:
\(a)\sqrt{x^2+2x+4}\ge x-2\\ b)x=\sqrt{x-\frac{1}{x}}+\sqrt{x+\frac{1}{x}}\\ c)\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2\sqrt{2x-5}}\\ d)x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ e)\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
giải phương trình
a)\(\sqrt{2x+1}+\sqrt{x+4}+x=3\)
b)\(\sqrt{x^2+3}+\sqrt{2x^2-1}=\sqrt{3x+6}\)
c) \(\left|x\right|+\sqrt{2x^2+2x+1}+\sqrt{x^2-6x+9}=2x+1\)
rút gọn:
B=\(1+\left(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}\right).\frac{x-\sqrt{x}}{2\sqrt{x}-1}\)
