a: \(=15x^4-12x^3+9x^2\)

c: \(=5x^3-15x^2-4x^2+12x\)

\(=5x^3-19x^2+12x\)

Đáp án:

 a.3x³−5x²+7xa.3x³−5x²+7x

b.−4x²y−10x²y+2xyb.−4x²y−10x²y+2xy

c.−x³+2x²+29x+20c.−x³+2x²+29x+20

d.2x⁴−3x³+2x²+3x−4d.2x⁴−3x³+2x²+3x−4

e.x²−4y²e.x²−4y²

h.2x²−6x+13h.2x²−6x+13

g.3xy⁴−12y²+2x²yg.3xy⁴−12y²+2x²y 

f.−2x²y³+y−3f.−2x²y³+y−3

Giải thích các bước giải:

 a.3x.(x²−5x+7)a.3x.(x²−5x+7)

=3x³−5x²+7x=3x³−5x²+7x

b.−2xy.(2x³+5x−1)b.−2xy.(2x³+5x−1)

=−4x⁴y−10xy²+2xy=−4x⁴y−10xy²+2xy

c.(x+4).(−x²+6x+5)c.(x+4).(−x²+6x+5)

=−x³+6x²+5x−4x²+24x+20=−x³+6x²+5x−4x²+24x+20

=−x³+2x²+29x+20=−x³+2x²+29x+20

d.(x²−1).(2x²−3x+4)d.(x²−1).(2x²−3x+4)

=2x⁴−3x³+4x²−2x²+3x−4=2x⁴−3x³+4x²−2x²+3x−4

=2x⁴−3x³+2x2+3x−4=2x⁴−3x³+2x2+3x−4

$e.(x+2y).(x-2y)$

=x²−(2y)²=x²−(2y)²

=x²−4y²=x²−4y²

h.(3x−1)²−7(x²+2)h.(3x−1)²−7(x²+2)

=9x²−6x+1−7x²−14=9x²−6x+1−7x²−14

=2x²−6x+13=2x²−6x+13

g.(6x²g.(6x²y⁵−xy³+4x³y²):2xy−xy³+4x³y²):2xy

=3xy⁴−12y²+2x²y=3xy⁴−12y²+2x²y 

f.(−12x³y⁴+6xy²−18xy):6xyf.(−12x³y⁴+6xy²−18xy):6xy

=−2x³y³+y−3

\(a)-2y(5xy+3x^2)\\=-10xy^2-6x^2y\\b)(2x+y)(4x-y)\\=8x^2-2xy+4xy-y^2\\=8x^2+2xy-y^2\\c)(x-1^2)-(x+1^2)\\=(x-1)-(x+1)\\=x-1-x-1\\=-2(bạn.xem.lại.đề.phần.này)\\d)(xem.lại.đề)\)

`#3107.101107`

a)

`A + B =` \(x^2+5xy-3y^2\)\(+ 2x^2-3xy+11y^2\)

`= (x^2 + 2x^2) + (5xy - 3xy) + (-3y^2 + 11y^2)`

`= 3x^2 + 2xy + 8y^2`

b)

\((9x^3y^2-12x^2y+15xy) \div (3xy)\)

`= 9x^3y^2 \div 3xy - 12x^2y \div 3xy + 15xy \div 3xy`

`= 3x^2y - 4x + 5`

a: \(-2y\left(5xy+3x^2\right)=-10xy^2-6x^2y\)

b: \(\left(2x+y\right)\left(4x-y\right)\)

\(=8x^2-2xy+4xy-y^2\)

\(=8x^2+2xy-y^2\)

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

a) \(x^2y\left(5xy-2x^2y-y^2\right)\)

\(=5x^3y^2-2x^4y^2-x^2y^3\)

b) \(\left(x-2y\right)\left(2x^3+4xy\right)\)

\(=2x^4+4x^2y-4x^3y-8xy^2\)

\(6x^2y+5xy-4xy\\ =6x^2y+\left[\left(5+4\right)xy\right]\\ =6x^2y+\left(1xy\right)\\=6x^2y+xy\)

`6x^{2}y+5xy-4xy`

`=6x^{2}y+[(5-4)xy]`

`=6x^{2}y+(1xy)`

`=6x^{2}y+xy`

`@Shả`

\(=2y\cdot\dfrac{5y}{4}\cdot\dfrac{16}{25x^2y^3}=\dfrac{8}{5x^2y}\)