Giải phương trình sau "
a, \(\sqrt{2x-1}=\sqrt{x^2+2x-5}.\)
b, \(\sqrt{x\left(x^3-3x+1\right)}=\sqrt{x\left(x^3-x\right)}\)
c, \(\sqrt{4x+1}-\sqrt{3x+4}=\sqrt{x-2}\)
d, \(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
e, \(\sqrt{x+2}-\sqrt{2x-3}=\sqrt{3x-5}\)
f, \(\sqrt{x\left(x-1\right)+\sqrt{x\left(2x-1\right)}=x}\)
g, \(\sqrt{x+1}+\sqrt{x-1}=2\)
h, \(\sqrt{2x-3}-\sqrt{4x+3}=-3\)
Mn giúp với cần gấp bài toan nâng co giải dc thì tick nhiều
a, \(\sqrt{x^2+2x-5}\)= \(\sqrt{2x-1}\)( x \(\ge\frac{1}{2}\))
\(\Leftrightarrow x^2+2x-5=2x-1\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-2\left(ktm\right)\end{cases}}\)
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b, \(\sqrt{x\left(x^3-3x+1\right)}\)\(=\sqrt{x\left(x^3-x\right)}\)\(\left(x\ge1\right)\)
\(\Leftrightarrow x\left(x^3-3x+1\right)\)= \(x\left(x^3-1\right)\)
\(\Leftrightarrow\)x( x3 - 3x + 1 ) - x ( x3 - 1 ) = 0
\(\Leftrightarrow\)x ( x3 - 3x + 1 - x3 + 1 ) = 0
\(\Leftrightarrow\)x( 2-3x ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2-3x=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{2}{3}\left(ktm\right)\end{cases}}\)
vậy pt vô nghiệm
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