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Bài 1:
a: \(=2x^2-3xy+5y^2\)
b: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
c: \(=\dfrac{6x^3+3x^2-10x^2-5x+4x+2}{2x+1}=3x^2-5x+2\)
c: \(=\dfrac{\left(x+3\right)^2-y^2}{x+y+3}=x+3-y\)
a) 2x.(3x2 – 5x + 3)
=2x3-10x2+6x
b(-2x-1).( x2 + 5x – 3 ) – (x-1)3
=-2x3 - 10x2 + 6x - x2 - 5x + 3 - x3 + 3x2 - 3x + 1
= -3x3 - 8x2 - 2x + 4
d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2
=2x2-3xy+5y2
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
\(a,=10x^3-5x^2+5x\\ b,=x^3+27\\ c,=\dfrac{5}{2}xy-1-\dfrac{1}{2}y\\ d,=\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)
a: =>3x+10-2x=0
hay x=-10
c: \(\Leftrightarrow3x^2-3x^2+6x=36\)
=>6x=36
hay x=6
a) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-4x^2y+2xy^2-4xy^2+2xy^2-y^3\)
\(=8x^3-8x^2y+4xy^2-y^3\)
b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)
\(=2x^2-3xy+5y^2\)