Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=-3x^3y^3-3x^2y^2+2x^2y\)
b: \(=6x^2+12x-2x-4\)
\(=6x^2+10x-4\)
c: \(=6x^3y^3+10x^2y^2-2x^2y\)
d: \(=2x^2-3x-2x+3\)
\(=2x^2-5x+3\)
Bài giải:
a) (-2x5 + 3x2 – 4x3) : 2x2 = (- 2222)x5 – 2 + 3232x2 – 2 + (-4242)x3 – 2 = - x3 + 3232 – 2x.
b) (x3 – 2x2y + 3xy2) : (- 1212x) = (x3 : -1212x) + (-2x2y : -1212x) + (3xy2 : -1212x)
= -2x2 + 4xy – 6y2
c)(3x2y2 + 6x2y3 – 12xy) : 3xy = (3x2y2 : 3xy) + (6x2y2 : 3xy) + (-12xy : 3xy)
= xy + 2xy2 – 4.
a) (-2x5+3x2-4x3) : 2x2
= (-2x5:2x2)-(4x3:2x2)+(3x2:2x2)
= -x3-2x+\(\dfrac{3}{2}\)
b) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)
= \(\left(x^3:\dfrac{-1}{2}x\right)+\left(-2x^2y:\dfrac{-1}{2}x\right)+\left(3xy^2:\dfrac{-1}{2}x\right)\)
= \(-2x^2+4xy-6y^2\)
c) \(\left(3x^2y^2+6x^2y^3-12xy\right):3xy\)
= \(\left(6x^2y^3:3xy\right)+\left(3x^2y^2:3xy\right)+\left(-12xy:3xy\right)\)
= \(xy^2+xy-4\)
cau a : (3x^2y-6xy+9x)(-4/3xy)
=-4/3xy.3x^2y+4/3xy.6xy-4/3xy.9x
=-4x+8-8y
cau b : (1/3x+2y)(1/9x^2-2/3xy+4y^2)
=(1/3)^3-2/9x^2y+8y^3+4/3xy^2+2/9x^2y-4/3xy^2+8y^3
=(1/3)^3 + (2y)^3x-2
cau c : (x-2)(x^2-5x+1)+x(x^2+11)
=x^3-5x^2+x-2x^2+10x-2+x^3+11x
=2x^3-7x^2+22x-2
cau d := x^3 + 6xy^2 -27y^3
cau e := x^3 + 3x^2 -5x - 3x^2y - 9xy = 15y
cau f := x^2-2x+2x -4-2x-1
= x(x-2)-5
a) x2(5x3 – x - \(\dfrac{1}{2}\) )= x2. 5x3 + x2 . (-x) + x2 . (-\(\dfrac{1}{2}\))
= 5x5 – x3 – \(\dfrac{1}{2}\)x2
b) (3xy – x2 + y)\(\dfrac{2}{3}\)x2y = \(\dfrac{2}{3}\)x2y . 3xy + \(\dfrac{2}{3}\)x2y . (- x2) + \(\dfrac{2}{3}\)x2y . y
= 2x3y2 – \(\dfrac{2}{3}\)x4y + \(\dfrac{2}{3}\)x2y2
c) (4x3– 5xy + 2x)(- \(\dfrac{1}{2}\)xy) = - \(\dfrac{1}{2}\)xy . 4x3 + (- \(\dfrac{1}{2}\)xy) . (-5xy) + (- \(\dfrac{1}{2}\)xy) . 2x
= -2x4y + \(\dfrac{5}{2}\)x2y2 - x2y.
a) x2 (5x3 - x - \(\dfrac{1}{2}\))
= 5x5 - x3 - \(\dfrac{1}{2}\)x2
b) (3xy - x2 + y) \(\dfrac{2}{3}\)x2y
= 2x3y2 - \(\dfrac{2}{3}\)x4y + \(\dfrac{2}{3}\)x2y2
c) (4x3 - 5xy +2x) (-\(\dfrac{1}{2}\)xy)
= -2x4y + \(\dfrac{5}{2}\)x2y2 - x2y
a: \(=2x^2-x+5\)
b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c: \(=-x^3+\dfrac{3}{2}-2x\)
d: \(=-2x^2+4xy-6y^2\)
e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
1a) (x - 2y) (x2 - 2xy + y2)
= (x - 2y) (x - y)2
= x2 - xy - 2xy + 2y2
= (x2 - xy) - (2xy - 2y2)
= x (x - y) - 2y (x - y)
= (x - y) (x - 2y)
2a) x (x - 3) - y (3 - x)
= x (x - 3) + y (x - 3)
= (x - 3) (x + y)
b) 3x2 - 5x - 3xy + 5y
= (3x2 - 3xy) - (5x - 5y)
= 3x (x - y) - 5 (x - y)
= (x - y) (3x - 5)
3) 12x (3 - 4x) + 7 (4x - 3) = 0
12x (3 - 4x) - 7 (3 - 4x) = 0
(3 - 4x) (12x - 7) = 0
=> 3 - 4x = 0 hoặc 12x - 7 = 0
* 3 - 4x = 0 => x = \(\frac{3}{4}\)
* 12x - 7 = 0 => x = \(\frac{7}{12}\)
Vậy x =\(\frac{3}{4}\)hoặc x =\(\frac{7}{12}\)
a) \((2x^2−3x)(5x^2−2x+1)\)
\(=2x^2.5x^2−2x^2.2x+2x^2−3x.5x^2+3x.2x−3x\)
\(=10x^4−4x^3+2x^2−15x^3+6x^2−3x\)
\(=10x^4−19x^3+8x^2−3x\)
b) \((x−2y)(3xy+5y^2+x)\)
\(=x.3xy+x.5y^2+x.x−2y.3xy−2y.5y^2−2y.x\)
\(=3x^2y+5xy^2+x^2−6xy^2−10y^3−2xy\)
\(=3x^2y−xy^2−2xy+x^2−10y^3\)