Question

[Zi Heo]

giúp e vs mn ơi

Answer 1

[Thanh Hoàng Thanh]

\(a.\left|x-2\right|+3=x.\\ \Leftrightarrow\left|x-2\right|=x-3.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3>0.\\\left(\left|x-2\right|\right)^2=\left(x-3\right)^2.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\x^2-4x+4=x^2-6x+9.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\2x=5.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>3.\\x=\dfrac{5}{2}.\end{matrix}\right.\) \(\Leftrightarrow x\in\phi.\)

\(b.\left(3x-4\right)\left(2x-5\right)=\left(3x-4\right)\left(x+2\right).\\ \Leftrightarrow\left(3x-4\right)\left(2x-5-x-2\right)=0.\\ \Leftrightarrow\left(3x-4\right)\left(x-7\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}.\\x=7.\end{matrix}\right.\)

\(\dfrac{x}{x-2}+\dfrac{x-1}{x}=2.\left(x\ne2;0\right).\\ \Leftrightarrow\dfrac{x^2+\left(x-1\right)\left(x-2\right)-2x\left(x-2\right)}{x\left(x-2\right)}=0.\\ \Rightarrow x^2+x^2-2x-x+2-2x^2+4x=0.\\ \Leftrightarrow x=-2\left(TM\right).\)

\(d.\dfrac{x-2}{2}-\dfrac{x+5}{3}=1-\dfrac{x-2}{4}.\\ \Leftrightarrow\dfrac{6x-12-4x-20-12+3x-6}{12}=0.\\ \Rightarrow5x=50.\\ \Leftrightarrow x=10.\)