\(\left(2x^3y-0,5x^2\right)^3\)

b,...">

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20 tháng 6 2018

a) \(\left(2x^3y-0,5x^2\right)^3\)

\(=\left(2x^3y\right)^3-3\left(2x^3y\right)^20,5x^2+3.2x^3y\left(0,5x^2\right)^2-\left(0,5x^2\right)^3\)

\(=8x^9y^3-6x^8y^2+1,5x^7y-0,125x^6\)

b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=x^3-\left(3y\right)^3\)

\(=x^3-27y^3\)

c) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)

\(=x^3-3^3\)

\(=x^3-27.\)

19 tháng 6 2018

a,\(\left(2x^3y-0,5x^2\right)^3=\left(2x^3y\right)^3-3.\left(2x^3y\right)^2.\left(0,5x^2\right)+3.\left(0,5x^2\right)^2.\left(2x^3y\right)-\left(0,5x^2\right)^3\)

\(=8x^9y^3-6x^8y^2+\frac{3}{2}x^7y-\frac{1}{8}x^6\)

b,\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)

\(=x^3-\left(3y\right)^3=x^3-27y^3\)

\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)

\(=\left(x^2\right)^3-3^3=x^6-27\)

a) Ta có: \(\left(x-3\right)^3\)

\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)

\(=x^3-9x^2+27x^2-27\)

b) Ta có: \(\left(2x-3\right)^3\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)

\(=8x^3-36x^2+54x-27\)

c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)

\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)

\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)

d) Ta có: \(\left(x^2-2\right)^3\)

\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)

\(=x^6-6x^4+12x^2-8\)

e) Ta có: \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)

\(=8x^3-24x^2y+36xy^2-27y^3\)

f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)

\(=\left(\frac{1}{2}x\right)^3-3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2-\left(y^2\right)^3\)

\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)

a) Ta có: \(\left(x+1\right)^3\)

\(=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)

\(=x^3+3x^2+3x+1\)

b) Ta có: \(\left(2x+3\right)^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=8x^3+3\cdot4x^2\cdot3+27\cdot2x+27\)

\(=8x^3+36x^2+54x+27\)

c) Ta có: \(\left(x+\frac{1}{2}\right)^3\)

\(=x^3+2\cdot x^2\cdot\frac{1}{2}+2\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)

\(=x^3+x^2+\frac{1}{2}x+\frac{1}{8}\)

d) Ta có: \(\left(x^2+2\right)^3\)

\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2+2^3\)

\(=x^6+6x^4+12x^2+8\)

e) Ta có: \(\left(2x+3y\right)^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2+\left(3y\right)^3\)

\(=8x^3+36x^2y+54xy^2+27y^3\)

f) Ta có: \(\left(\frac{1}{2}x+y^2\right)^3\)

\(=\left(\frac{1}{2}x\right)^3+3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)

\(=\frac{1}{8}x^3+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)

17 tháng 11 2022

a: \(=4x^2-25-4x^2+12x-9-12x=-34\)

b: \(=8y^3-12y^2+6y-1-2y\left(4y^2-12y+9\right)-12y^2+12y\)

\(=8y^3-24y^2+18y-1-8y^3+24y^2-18y=-1\)

c: \(=x^3+27-x^3-20=7\)

d: \(=3y\left(9y^2+12y+4\right)-27y^3+1-36y^2-12y-1\)

\(=27y^3+36y^2+12y-27y^3-36y^2-12y\)

=0

a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)\)

\(=\left(x-3\right)\left(x^2+x\cdot3+3^2\right)\)

\(=x^3-3^3=x^3-27\)

b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x-2\right)\left(x^2+x\cdot2+2^2\right)\)

\(=x^3-2^3=x^3-8\)

c) Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)\)

\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)

\(=x^3+4^3=x^3+64\)

d) Ta có: \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=\left(x-3y\right)\left[x^2+x\cdot3y+\left(3y\right)^2\right]\)

\(=x^3-\left(3y\right)^3=x^3-27y^3\)

e) Ta có: \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\)

\(=\left(x^2-\frac{1}{3}\right)\left[\left(x^2\right)^2+x^2\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3\)

\(=x^6-\frac{1}{27}\)

f) Ta có: \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)

\(=\left(\frac{1}{3}x+2y\right)\left[\left(\frac{1}{3}x\right)^2-\frac{1}{3}x\cdot2y+\left(2y\right)^2\right]\)

\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\frac{1}{27}x^3+8y^3\)

27 tháng 9 2019

nhân ra hết là đc

27 tháng 9 2019

kệ chẳng quan tâm

21 tháng 10 2020

cau a : (3x^2y-6xy+9x)(-4/3xy)

           =-4/3xy.3x^2y+4/3xy.6xy-4/3xy.9x

           =-4x+8-8y

cau b : (1/3x+2y)(1/9x^2-2/3xy+4y^2)

            =(1/3)^3-2/9x^2y+8y^3+4/3xy^2+2/9x^2y-4/3xy^2+8y^3

             =(1/3)^3 + (2y)^3x-2

cau c :  (x-2)(x^2-5x+1)+x(x^2+11)

            =x^3-5x^2+x-2x^2+10x-2+x^3+11x

            =2x^3-7x^2+22x-2

cau d := x^3 + 6xy^2 -27y^3

cau e := x^3 + 3x^2 -5x - 3x^2y - 9xy = 15y

cau f := x^2-2x+2x -4-2x-1

          = x(x-2)-5

21 tháng 10 2020

cau e la + 15y ko phai =15y

19 tháng 6 2018

a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)

\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)

b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36y^2x^2-8y^3\)

c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)