Thực hiện phép tính :
a) \(\left(-6x^3+7x^2-5x+2\right):\left(-2x+1\right)\)
b) \(\left(x^4-x^2+2x-1\right):\left(x^2+x-1\right)\)
K
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1 tháng 7 2023
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 7 2023
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
25 tháng 9 2020
b) Ta có: \(\left(x^4-x^2+2x-1\right):\left(x^2+x-1\right)\)
\(=\frac{x^4-\left(x^2-2x+1\right)}{x^2+x-1}\)
\(=\frac{x^4-\left(x-1\right)^2}{x^2+x-1}\)
\(=\frac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{x^2+x-1}\)
\(=x^2-x+1\)
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
2
14 tháng 12 2022
`a)3x(2x^2-3x+4)`
`=6x^3-9x^2+12x`
______________________________________________
`b)(x+3)^2+(3x-2)(x+4)`
`=x^2+6x+9+3x^2+12x-2x-8`
`=4x^2+16x+1`
______________________________________________
`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]` `ĐK: x \ne +-1`
`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`
`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`
`=[2x^2-2]/[x^2-1]`
`=2`
19 tháng 8 2020
a)\(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=2x^2\left(5x^2-2x+1\right)-3x\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
KN
19 tháng 8 2020
a. \(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
b. \(\left(2x^4-x^3+3x^2\right):\left(\frac{1}{3}x^2\right)\)
\(=\left(2x^4-x^3+3x^2\right).\frac{3}{x^2}\)
\(=0,6x^2-3x+0,9\)
27 tháng 9 2020
a) \(\left(2x-3\right)\left(x^2-2x+1\right)+2\left(2-x\right)^3\)
\(=2x\left(x^2-2x+1\right)-3\left(x^2-2x+1\right)+2\left(2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+2\left(8-12x+6x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+16-24x+12x^2-2x^3\)
\(=\left(2x^3-2x^3\right)+\left(-4x^2-3x^2+12x^2\right)+\left(2x+6x-24x\right)+\left(-3+16\right)\)
\(=5x^2-16x+13\)
b)
Vậy \(\left(2x^3-7x^2+2x+3\right):\left(x^2-4x+3\right)=2x+1\)
Câu b thêm dấu " - " ở chỗ 2x3 - 7x2 + 2x +3 và 2x3 - 8x2 + 6x nhé :)))
16 tháng 5 2022
a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
29 tháng 7 2021
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
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