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a: Ta có: \(x\left(2-3x\right)+\left(3x^3-x^2\right):x\)
\(=2x-3x^2+3x^2-x\)
=x
b: Ta có: \(2x\left(x-3y\right)-\left(8x^3y-12x^2y^2\right):2xy\)
\(=2x^2-6xy-4x^2+6xy\)
\(=-2x^2\)
\(1,\\ a,\dfrac{x^2}{x+1}+\dfrac{x}{x+1}=\dfrac{x^2+x}{x+1}=\dfrac{x\left(x+1\right)}{x+1}=x\)
\(b,\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right):\dfrac{x+y}{2x}=\left(\dfrac{4xy}{2\left(x-y\right)\left(x+y\right)}+\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\right).\dfrac{2x}{x+y}=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}.\dfrac{2x}{x+y}=\dfrac{2x\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{2x\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)^2}=\dfrac{x}{x-y}\)
\(a,=-15x^3+10x^4+20x^2\\ b,=2x^3+2x^2+4x-x^2-x-2=2x^3+x^2+3x-2\)
\(a)\left(x+3y\right)\left(x-2y\right)\\ =x^3-2xy+3xy-6y^2\\ =x^2+xy-6y^2\\ b)\left(2x-y\right)\left(y-5x\right)\\ = 2xy-10x^2-y^2+5xy\\ =7xy-10x^2-y^2\\ c)\left(2x-5y\right)\left(y^2-2xy\right)\\ =2xy^2-4x^2y-5y^3+10xy^2\\ =12xy^2-4x^2y-5y^2\\ d)\left(x-y\right)\left(x^2-xy-y^2\right)\\ =x^3-x^2y-xy^2-x^2y+xy^2+y^3\\ =x^3-2x^2y+y^3\)
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
Bài 2:
a: =x(x^2-25)
=x(x-5)(x+5)
b: =x(x-2y)+3(x-2y)
=(x-2y)(x+3)
c: =(2x-3)(4x^2+6x+9)+2x(2x-3)
=(2x-3)(4x^2+8x+9)
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
\(x\left(2-3x\right)+\left(3x^2-x^2\right):x\)
\(=2x-3x^2+3x^2-x\)
\(=x\)
\(2x\left(x-3y\right)-\left(8x^3y-12x^2y^2\right):2xy\)
\(=2x^2-6xy-4x^2+6xy\)
\(=-2x^2\)