Thực hiện phép chia
a). (3x2y2 + 6x2y3 - 12xy) : 3xy
b). (2x3 + 5x2 - 2x + 3) : (2x2 - x + 1)
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10 tháng 1 2022
a) \(=6x^3+8x^2+2x-6x^3=8x^2+2x\)
b) \(=\left[3xy\left(xy+2xy^2-4\right)\right]:3xy=xy+2xy^2-4\)
c) \(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x+2}-\dfrac{5}{x-2}=\dfrac{10x+3\left(x-2\right)-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)
DT
10 tháng 1 2022
a, \(=6x^3+12x^2+2x-6x^3\\=12x^2+2x\)
b,
\(=xy+2xy^2-4\)
c,
\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)
\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)
25 tháng 10 2021
Câu 3:
a: 2x-8=4
nên 2x=12
hay x=6
b: 7x-3x=2x+7
\(\Leftrightarrow4x-2x=7\)
hay \(x=\dfrac{7}{2}\)
25 tháng 10 2021
Câu 1:
a: \(5x\left(3x-4\right)=15x^2-20x\)
b: \(\left(x+5\right)\left(x-5\right)=x^2-25\)
YN
2 tháng 10 2021
a) \(\left(5x^2-2x+1\right)\left(2x^2-3x\right)\)
\(=10x^4-15x^3-4x^3+6x^2+2x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
1 tháng 10 2021
a)(5x2-2x+1).(2x2-3x)
=10x4-4x3+2x2-15x3+6x2-3x
=10x4-19x3+8x2-3x
b)(18x4y3-6x2y3+12x3y4z):6x2y3
=(18x4y3:6x2y3)-(6x2y3:6x2y3)+(12x3y4z:6x2y3)
=3x2y-xy+2xyz
7 tháng 12 2021
\(a,=2xy^3\\ b,=xy+3xy^2-4\\ c,=\left(x-4\right)\left(x+4\right):\left(x-4\right)=x+4\\ d,=\left(x+5\right)^2:\left(x+5\right)=x+5\)
27 tháng 7 2021
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)
\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)
\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)
\(=5x^2+4x+4\)
27 tháng 7 2021
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
3 tháng 6 2023
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: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
\(b,=3xy\left(xy+2xy^2-4\right):3xy\\ =xy+2xy^2-4\\ c,=\left[\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)\right]:\left(2x^2-x+1\right)\\ =\left[x\left(2x^2-x+1\right)+3\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\\ =\left[\left(x+3\right)\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\\ =x+3\)
b: \(=xy+2xy^2-4\)