a) Ta có: \(-5x^2+3x=0\)
\(\Leftrightarrow x\left(-5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{3}{5}\right\}\)
b) Ta có: \(1+\frac{x-1}{3}=\frac{2x+1}{6}-2\)
\(\Leftrightarrow1+\frac{x-1}{3}-\frac{2x+1}{6}+2=0\)
\(\Leftrightarrow3+\frac{x-1}{3}-\frac{2x+1}{6}=0\)
\(\Leftrightarrow\frac{18}{6}+\frac{2\left(x-1\right)}{6}-\frac{2x+1}{6}=0\)
\(\Leftrightarrow18+2x-2-2x-2=0\)
\(\Leftrightarrow14=0\)(vô lý)
Vậy: x∈∅
c) Ta có: 2-x=3(x+1)
⇔2-x=3x+3
⇔2-x-3x-3=0
⇔-4x-1=0
⇔-4x=1
hay \(x=\frac{-1}{4}\)
Vậy: \(x=\frac{-1}{4}\)
d) Ta có: 4x+7(x-2)=-9x+5
⇔4x+7x-14+9x-5=0
⇔20x-19=0
⇔20x=19
hay \(x=\frac{19}{20}\)
Vậy: \(x=\frac{19}{20}\)
e) Ta có: -4(x+3)=5(2x-9)
⇔-4x-12=10x-45
⇔-4x-12-10x+45=0
⇔-14x+33=0
⇔-14x=-33
hay \(x=\frac{33}{14}\)
Vậy: \(x=\frac{33}{14}\)
f) Ta có: \(\frac{2x-1}{3}-\frac{5x+2}{4}=2x\)
\(\Leftrightarrow\frac{4\left(2x-1\right)}{12}-\frac{3\left(5x+2\right)}{12}=\frac{24x}{12}\)
\(\Leftrightarrow4\left(2x-1\right)-3\left(5x+2\right)-24x=0\)
\(\Leftrightarrow8x-4-15x-6-24x=0\)
\(\Leftrightarrow-31x-10=0\)
\(\Leftrightarrow-31x=10\)
hay \(x=\frac{-10}{31}\)
Vậy: \(x=\frac{-10}{31}\)
