Làm tính chia:
\(\left(x^4-x^3+x^2+3x\right):\left(x^2-2x+1\right)\)
K
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KV
7 tháng 1 2021
a) (x + 2)(x2 + 3x + 1)
= x.x2 + x.3x + x.1 + 2.x2 + 2.3x + 2.1
= x3 + 3x2 + x + 2x2 + 6x + 2
= x3 + 5x2 + 7x + 2
b) (2x3 + 10x2 + 9x + 4) : (x + 4)
= (2x3 + 8x2 + 2x2 + 8x + x + 4) : (x + 4)
= [(2x3 + 8x2) + (2x2 + 8x) + (x + 4)] : (x + 4)
= [2x2(x + 4) + 2x(x + 4) + (x + 4)] : (x + 4)
= (x + 4)(2x2 + 2x + 1) : (x + 4)
= 2x2 + 2x + 1
KK
25 tháng 10 2017
x -1 2x -5x +x +3x-1 2 5 3 2 2x 3 2x -2x 5 3 -3x +x +3x-1 3 2 -3x -2 -3x +3x 3 2 -2x +3x-1 2 2 -2x +2 3x -3
16 tháng 5 2022
a: \(=\dfrac{6x^2+15x-2x-5}{2x+5}=3x-1\)
b: \(=\dfrac{x^2\left(x+3\right)+\left(x-3\right)}{x-3}=x^2+1\)
c: \(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}=2x^2+x+1\)
17 tháng 10 2017
a) \(\left( {6{x^3} - 7{x^2} - x + 2} \right):\left( {2x + 1} \right)\)
b) $(x^4-x^3+x^2+3x):(x^2-2x+3)$
c) \(\left( {{x^2} + {y^2} + 6x + 9} \right):\left( {x + y + 3} \right)\)
\(=\left( {{x^2} + 6x + 9 - {y^2}} \right)\left( {x + y + 3} \right)\)
\(=\left[ {\left( {{x^2} + 2x.3 + {3^2}} \right) - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left[ {{{\left( {x + 3} \right)}^2} - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left( {x + 3 - y} \right)\left( {x + 3 + y} \right):\left( {x + y + 3} \right)\)
$= x + 3 - y$
$= x - y + 3$
KN
26 tháng 10 2017
(6x3 - 7x2 - x + 2) : (2x + 1)
= (6x3 + 3x2 - 10x2 - 5x + 4x + 2) : (2x + 1)
= [(6x3 + 3x2) - (10x2 + 5x) + (4x + 2)] : (2x + 1)
= [3x2(2x + 1) - 5x(2x + 1) + 2(2x + 1)] : (2x + 1)
= (3x2 - 5x + 2)(2x + 1) : (2x + 1)
= 3x2 - 5x + 2
(x4 - x3 + x2 + 3x) : (x2 - 2x + 3)
= (x4 + x3 - 2x3 - 2x2 + 3x2 + 3x) : (x2 - 2x + 3)
= [(x4 + x3) - (2x3 + 2x2) + (3x2 + 3x)] : (x2 - 2x + 3)
= [x3(x + 1) - 2x2(x + 1) + 3x(x + 1)] : (x2 - 2x + 3)
= (x3 - 2x2 + 3x)(x + 1) : (x2 - 2x + 3)
= x(x2 - 2x + 3)(x + 1): (x2 - 2x + 3)
= x(x + 1)
= x2 + x
(x2 - y2 + 6x + 9) : (x + y + 3)
= [(x2 + 6x + 9) - y2] : (x + y + 3)
= [(x + 3)2 - y2] : (x + y + 3)
= (x + 3 + y)(x + 3 - y) : (x + y + 3)
= (x + y + 3)(x - y + 3) : (x + y + 3)
= x - y + 3
CHÚC BN HOK TỐT 
5 tháng 10 2021
b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow-4x+3+5x+2=0\)
\(\Leftrightarrow x=-5\)
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`
SG
2
20 tháng 4 2017
Ta có:\(\left(2x^4+x^3-3x^2+5x-2\right):\left(x^2-x+1\right)\)
= \(\left(2x^4-2x^3+2x^2+3x^3-3x^2+3x-2x^2+2x-2\right):\left(x^2-x+1\right)\)
=\(\left(\left(2x^4-2x^3+2x^2\right):\left(x^2-x+1\right)\right)\)+\(\left(\left(3x^3-3x^2+3x\right):\left(x^2-x+1\right)\right)\)+\(\left(\left(-2x^2+2x-2\right):\left(x^2-x+1\right)\right)\)
= \(2x^2.\left(x^2-x+1\right):\left(x^2-x+1\right)\)+\(3x.\left(x^2-x+1\right):\left(x^2-x+1\right)\)\(-2\left(x^2-x+1\right):\left(x^2-x+1\right)\)
= \(2x^2+3x-2\)
=(x.x^3-x.x^2-x.x+3.x)/(x.x-2x+1)
=x(x^3-x^2+x+3)/x(x-2+1)
=2x/x-1
(chắc lm như vầy)