3x^3-6x^2+3x
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a: Sửa đề: 6x^3
\(\dfrac{6x^3-4x^2+3x-2}{3x-2}=\dfrac{2x^2\left(3x-2\right)+3x-2}{3x-2}=2x^2+1\)
b: \(\dfrac{6x^3-3x^2+4x+2}{3x^2+2}\)
\(=\dfrac{6x^3+4x-3x^2-2+4}{3x^2+2}\)
\(=2x-1+\dfrac{4}{3x^2+2}\)
`1)` Yêu cầu là gì ạ?
`2)`
`P(x)-Q(x)=`\((6x^3-3x^2+5x-1)-(-6x^3+3x^2-2x+7)\)
`= 6x^3-3x^2+5x-1+6x^3-3x^2+2x-7`
`= (6x^3+6x^3)+(-3x^2-3x^2)+(5x+2x)+(-1-7)`
`= 12x^3-6x^2+7x-8`
`3)`
`(-3x^3+15x^2+81x):(-3x)`
`= (-3x^3) \div (-3x) + 15x^2 \div (-3x) + 81x \div (-3x)`
`= x^2-5x-27`
a: \(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
b: \(=27x^3-8-27x^3+6=-2\)
c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
\(3x^3-6x^2+3x\)
\(=3x\left(x^2-2x+1\right)\)
\(=3x\left(x-1\right)^2\)
\(3x^3-6x^2+3x\)
\(=3x\left(x^2-2x+1\right)\)
\(=3x\left(x-1\right)^2\)
Vậy \(3x^3-6x^2+3x=3x\left(x-1\right)^2\)