c: Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)\)

\(=6x^2-6x^2+4x-15x+10\)

=-11x+10

d: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

(1𝑦/3+3)^3

(𝑦/3+3)^3

(𝑦/3+3⋅3/3)^3

(𝑦+3⋅3/3)^3

(𝑦+9/3)^3

\(\left(\dfrac{1}{3}y+3\right)^3=\dfrac{1}{27}y^3+y^2+9y+27\)

-ab - ba

= -ab + (-ab)

= 2-ab (hai âm ab)

-ab - ba

= -ab + (-ab) 

= -2ab

=a³+2a²b+a²b+2ab²+ab²+b³

=(a³+2a²b+ab²)+(a²b+2ab²+b³)

=a(a²+2ab+b²)+b(a²+2ab+b²)

=(a+b)(a²+ab+ab+b²)

=(a+b)(a(a+b)+b(a+b))

=(a+b)(a+b)(a+b)=(a+b)³

=a³ + 2a² b+a2b+2ab2+ab3+b3

=(a3+2a2b+ab2)+(a2b+2ab2+b3)

=a(a2+2ab+b2)+b(a2+2ab+b3)

=(a+b)(a(a+b)+b(a-b)

=(a-b)(a+b)(a-b)=(a+b)2

nhớ đó, tau sẽ mách thầy

\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1}{\left(3y+3\right)^3}=\dfrac{1}{27y^3+81y^2+81y+27}\)

\(\left(\dfrac{1}{3y+3}\right)^3=\dfrac{1^3}{\left(3y+3\right)^3}=\dfrac{1}{27\left(y^3+3y^2+3y+1\right)}\)

\(\left(\dfrac{1}{3}y+3\right)^3=\dfrac{1}{27}y^3+y^2+9y+27\)

Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

\(\text{ 2x.(x-2)+(x+3).(1-2x)}\\ =\left(2x^2-4x\right)+\left(x-2x^2+3-6x\right)\\ =2x\left(x-2\right)+\left(-5-2x^2+3\right)\)

\(=8x^3-36x^2+54x-27+2x^2-8x^3-29=-34x^2+54x-56\)

\(=8x^3-36x^2+54x-27+2x^2-8x^3-29\)

\(=-34x^2+54x-56\)