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9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
a) \(\left(2x^3-x^2+5x\right):x\)
\(=\dfrac{2x^3-x^2+5x}{x}\)
\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)
\(=2x^2-x+5\)
b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)
\(=\dfrac{3x^4-2x^3+x^2}{-2x}\)
\(=\dfrac{2x\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)}{-2x}\)
\(=-\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)\)
\(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)
\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)
\(=\dfrac{2x^2\left(-x^3+\dfrac{3}{2}-2x\right)}{2x^2}\)
\(=-x^3-2x+\dfrac{3}{2}\)
d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)
\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)
\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)
\(=-\left(2x^2-4xy+6y^2\right)\)
\(=-2x^2+4xy-6y^2\)
e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)
\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)
\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)
\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)
\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)
\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)
x2 ( 2x3 – 4x + 3)
a: \(=15x^5-25x^4+15x^3\)
b: \(=2x^3+10x^2-8x-x^2-5x+4\)
\(=2x^3+9x^2-13x+4\)
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^4-10x^3+6x^2\)
c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)
d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
1. 2x(3x2 - 5x + 3) = 6x3 - 10x2 + 6x
2. \(-\dfrac{1}{2}x^2\left(2x^3-4x+3\right)=-x^5+2x^3+\dfrac{-3}{2}x^2\)
3. -2x(x2 + 5x - 3) = -2x3 - 10x2 + 6x
4. x(3x2 - 2x + 5) = 3x3 - 2x2 + 5x
5. 3xy2(2x - 4y + 3xy) = 6x2y2 - 12xy3 = 9x2y3