a) \(3x+2\left(5-x\right)=-11\)
\(3x+10-2x=-11\)
\(x=-21\)
b) \(3x^2-3x\left(-2+x\right)=36\)
\(3x^2+6x-3x^2=36\)
\(x=6\)
c) \(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(5x-2x^2+2x^2-2x=15\)
\(x=5\)
d) \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(36x^2-12x-36x^2+27x=30\)
\(x=2\)
e) \(3x\left(x+\dfrac{1}{2}\right)=3x^2-1\)
\(3x^2+\dfrac{3}{2}x-3x^2=-1\)
\(x=-\dfrac{2}{3}\)
f) \(2x^2-\left(2x^2-2x\right)+3=2x\)
\(2x^2-2x^2+2x+3-2x=0\)
3=0 ( vô lí ) => vô nghiệm
g) \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)
\(4x^4-6x^3-4x^4+6x^3-2x^2=0\)
\(x=0\)
a) Ta có: \(3x+2\left(5-x\right)=-11\)
\(\Leftrightarrow3x+10-2x=-11\)
\(\Leftrightarrow x=-21\)
b) Ta có: \(3x^2-3x\left(-2+x\right)=36\)
\(\Leftrightarrow3x^2+6x-3x^2=36\)
\(\Leftrightarrow6x=36\)
hay x=6
c) Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=15\)
\(\Leftrightarrow3x=15\)
hay x=5
d) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow15x=30\)
hay x=2
e) Ta có: \(3x\left(x+\dfrac{1}{2}\right)=3x^2-1\)
\(\Leftrightarrow3x^2+\dfrac{3}{2}x-3x^2=-1\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=-1\)
hay \(x=-\dfrac{2}{3}\)
f) Ta có: \(2x^2-\left(2x^2-2x\right)+3=2x\)
\(\Leftrightarrow2x^2-2x^2+2x+3-2x=0\)
\(\Leftrightarrow0x+3=0\)(Vô lý)
g) Ta có: \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)
\(\Leftrightarrow4x^4-6x^3-4x^4+6x^3-2x^2=0\)
\(\Leftrightarrow-2x^2=0\)
\(\Leftrightarrow x=0\)
