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\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
x3 + 2x2y + xy2
= x(x2 + 2xy + y2)
= x(x + y)2
x2 - xy - 4x + 4y
= x(x - y) - 4(x - y)
= (x - y)(x - 4)
1.
\(x^3+2x^2y+xy^2\\ =\left(x^3+x^2y\right)+\left(x^2y+xy^2\right)\\ =x^2\left(x+y\right)+xy\left(x+y\right)\\ =\left(x+y\right)\left(x^2+xy\right)\\ =\left(x+y\right)^2.x\)
\(x^2-xy-4x+4y\\ =\left(x^2-xy\right)-\left(4x-4y\right)\\ =x\left(x-y\right)-4\left(x-y\right)=\left(x-y\right)\left(x-4\right)\)
\(\dfrac{x-1}{x-2}+\dfrac{2x-3}{x-2}+\dfrac{x-4}{x-2}\\ =\dfrac{4x-8}{x-2}=4\)
Đây là cách hiện đại :
\(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-\left(2x^3-2x\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x^2+1\right)-2x\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(\left(x^2+1\right)-2x\right)\)
a,=\(x^4-x^3-x^3+x^2-x^2+x+x-1\)
cu hai so nhom 1 nhom roi dat thua so chung la xong
b,x^4+x^3+x^3+x^2+x^2+x+x+1
cu hai so lai nhom 1 nhom va dat thua so chung
\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)
= ( x3 + 2x2y + xy2 ) - 4y2
= x.( x2 + 2xy + y2 ) - 4y2
= x.( x + y )2 - 4y2
= x.[( x + y) - 4y]. [(x + y) + 4y]