Câu 1 :
a. \(3x-2=10\\ \Leftrightarrow3x=10+2\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\)
b. \(\dfrac{3x+2}{x-1}=1+\dfrac{x}{x-1}\\ ĐKXĐ:x\ne1\\ \Leftrightarrow\dfrac{3x+2}{x-1}=\dfrac{x-1}{x-1}+\dfrac{x}{x-1}\\ \Rightarrow3x+2=x-1+x\\ \Leftrightarrow3x-x-x=-1-2\\ \Leftrightarrow x=-3\left(nhận\right)\)
c. \(3x\left(x-1\right)-2\left(1-x\right)=0\\ \Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\x=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)
a. 3x-2=10=>3x=12=>x=4
b.\(\dfrac{x-1+2x+3}{x-1}=1+\dfrac{x}{x-1}=1+\dfrac{2x+3}{x-1}=1+\dfrac{x}{x-1}=>x=2x+3=>x=\left(-3\right)\)
c.3x(x-1)-2(1-x)=0
\(3x^2-3x-2-2x=0\)
\(3x^2-5x-2=0\)
\(x\left(3x-5\right)=2\)
\(=>\left[{}\begin{matrix}x=1\left(ktm\right)\\x=2\left(tm\right)\\x=-1\left(ktm\right)\\x=\left(-2\right)\left(ktm\right)\end{matrix}\right.\)
=>x=2
