a,

$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$

b,

$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$

c,

$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$

d,

$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$

$=3x^2+x$

e,

$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$

f,

$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$

$=\frac{23}{4}xy^2$

a) \(x^2y\left(5xy-2x^2y-y^2\right)\)

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

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

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

Thực hiện phép tính: 5xy + 6xy - 4xy

\(5xy+6xy-4xy\\ =\left(5+6-4\right)xy\\ =7xy\)

\(2xy+4xy-5xy=6xy-5xy=xy\)

\(x\left(3+2xy\right)=3x+2x^2y\)

`a)[3x+2]/[x^2]:[6x+4]/[2x^2]`       

`=[3x+2]/[x^2].[2x^2]/[2(3x+2)]`

`=1`

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`b)[4xy]/[x+y]:[6x^2y^3]/[x^2-y]`         

`=[4xy]/[x+y].[(x-y)(x+y)]/[6xy.xy^2]`

`=[2(x-y)]/[3xy^2]=[2x-2y]/[3xy^2]`

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

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

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

Answer:

\(3x^2.\left(2x^3-x+5\right)\)

\(=3x^2.2x^3+3x^2.(-x)+3x^2.5\)

\(=6x^5-3x^3+15x^2\)

\((4xy+3y-5x).x^2y\)

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

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