(3x - 1)³ = (3x)³ - 3 . (3x)² . 1 + 3 . 3x . 1^{2 - 13} = 27x³ - 27x² + 9x - 1

Ta có: \(\left(3x-1\right)^3\)

\(=\left(3x\right)^3-3\cdot\left(3x\right)^2+3\cdot3x-1\)

\(=27x^3-27x+9x-1\)

Em cần làm gì với đa thức này?

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

\(=\left(\dfrac{2x}{y}\right)^3-3\cdot\dfrac{4x^2}{y^2}\cdot\dfrac{3y}{x}+3\cdot\dfrac{2x}{y}\cdot\dfrac{9y^2}{x^2}-\left(\dfrac{3y}{x}\right)^3\)

\(=\dfrac{8x^3}{y^3}-\dfrac{36x}{y^2}+\dfrac{54y}{x}-\dfrac{27y^3}{x^3}\)

\(A=\left(2x-3y\right)\left(5x-2y\right)=10x^2-4xy-15xy+6y^2=10x^2-19xy+6y^2\)

   (2\(x\) + 3y)²

= (2\(x\) + 3y).(2\(x\) + 3y)

= 4\(x\)² + 6\(xy\) + 6\(xy\) + 9y²

= 4\(x^2\) + 12\(xy\) + 9y²

\(4x^2+12xy+9y^2\)

4�2x2 + 12$xy$ + 9y2

Trả lời:

a, \(\left(\frac{1}{2}x-3\right)^2=\left(\frac{1}{2}x\right)^2-2.\frac{1}{2}x.3+3^2=\frac{1}{4}x^2-3x+9\)

b, \(\left(2x^2+3y\right)^2=\left(2x^2\right)^2+2.2x^2.3y+\left(3y\right)^2=4x^4+12x^2y+9y^2\)

c, \(\left(3x-7y\right)^2=\left(3x\right)^2-2.3x.7y+\left(7y\right)^2=9x^2-42xy+49y^2\)

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

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

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

a. (2x+3y)²= (2x)²+2.2x.3y+(3y)²

=4x²+12xy+9y²

b. 2(\(\dfrac{1}{2}\)x²+y)(x²-2y)

=(x²+2y)(x²-2y)

=x⁴-4y²

c, (x+y+z)²= [(x+y)+z]²

=(x+y)²+2(x+y)z+z²

=x²+2xy+y²+2xz+2yz+z²

=x²+y²+z²+2xy+2yz+2xz

a) \(\left(2x+1\right)^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)

\(=8x^3+12x^2+6x+1\)

b) \(\left(x-3\right)^3\)

\(=x^3-3.x^2.3+3.x.3^2-3^3\)

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

Bài 2: 

a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)

b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)

c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)

=(5x³)³-3(5x³)²(2y²)+3(5x³)(2y²)²-(2y²)³

=125x⁹-150x⁶y²+60x³y⁴-8y⁶