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a) 2x. (x2 – 7x -3)
= 2x3- 14x2- 6x
b) ( -2x3 + y2 -7xy). 4xy2
= -8x4y2+ 4xy4- 28x2y3
c)(-5x3).(2x2+3x-5)
= -10x5-15x4+25x3
d) (2x2 - xy+ y2).(-3x3)
=-6x5+ 3x4y -3x3y2
e)(x2 -2x+3). (x-4)
=x3-2x2+3x -4x2+8x-12
=x3-6x2+11x-12
f) ( 2x3 -3x -1). (5x+2)
=10x4-15x2-5x +4x3-6x-2
=10x4+4x3-15x2-11x-2
a: \(=\dfrac{\left(x^2+5\right)\left(x^2-5\right)+2x\left(x^2+5\right)}{x^2+5}=x^2+2x-5\)
b: \(=\dfrac{x^3-2x^2-x^2+2x+3x-6}{x-2}=x^2-x+3\)
a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
Lời giải:
$2x^3-x^2+4=2x^2(x+1)-3x(x+1)+3(x+1)+1$
$=(x+1)(2x^2-3x+3)+1$
Với $x$ nguyên, để $2x^3-x^2+4\vdots x+1$ thì $1\vdots x+1$
$\Rightarrow x+1\in \text{Ư(1)}$
$\Rightarrow x+1\in\left\{\pm 1\right\}$
$\Rightarrow x\in\left\{-2; 0\right\}$
a) x(3 - x) + (x + 1)(x - 1)
= 3x - x2 + x2 - x + x - 1
= 3x - 1
a. 5x + 3(x2 - x - 1)
= 5x + 3x2 - 3x - 3
= 3x2 + 5x - 3x - 3
= 3x2 + 2x - 3
b. (5 - x)(5 + x) - (2x - 1)2
25 - x2 - (4x2 - 4x + 1)
= 25 - x2 - 4x2 + 4x - 1
= 25 - 1 - x2 - 4x2 + 4x
= 24 - 5x2 + 4x