a, \(3\left(2x-1\right)-3x\left(-x+2\right)=5x-\left(1-3x\right)\cdot x\\ 6x-3+3x^2-6x=5x-x+3x^2\\ 3x^2-3=4x+3x^2\\ 3x^2-3x^2=4x+3\\ 4x+3=0\\ 4x=-3\\ x=\frac{-3}{4}\)
Vậy \(x=\frac{-3}{4}\)
b, \(x-\frac{x-3}{4}=3-\frac{x-3}{12}\\ \frac{4x-x-3}{4}=\frac{36-x-3}{12}\\ \frac{3x-3}{4}=\frac{33-x}{12}\\ \Rightarrow12\left(3x-3\right)=4\left(33-x\right)\\ 36x-36=132-4x\\ 36x+4x=132+36\\ 40x=168\\ x=\frac{168}{40}=\frac{21}{5}\)
Vậy \(x=\frac{21}{5}\)
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