Question

Tính theo kiểu hằng đẳng thức mở rộng

a) \(\left(x+2y\right)^3\)

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

c)\(\left(x^2+x+1\right).\left(x-1\right)\)

d) \(\left(4x^2-2x+1\right).\left(2x+1\right)\)

Answer 1

d) \(\left(4x^2-2x+1\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left[\left(2x\right)^2-2x.1+1^2\right]\)

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

\(=8x^3+1\)

Answer 2

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

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

b) \(\left(2x-y\right)^3=\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y^2-y^3\)

\(=8x^3-12x^2y+6xy^2-y^3\)

c) \(\left(x^2+x+1\right).\left(x-1\right)=x^3-x^2+x^2-x+x-1\)

\(=\left(x^3-1\right)\)

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d) \(\left(4x^2-2x+1\right)\left(2x+1\right)=\left(2x+1\right)\left(4x^2-2x+1\right)\)

\(=\left(2x+1\right)\left(\left(2x\right)^2-2x.1+1^2\right)\)

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

\(=8x^3+1\)

Answer 3

a) \(\left(x+2y\right)^3\)

\(=x^3+3.x^2.2y+3.x.\left(2y\right)^2+\left(2y\right)^3\)

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

Answer 4

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

\(=\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y^2-y^3\)

\(=8x^3-12x^2y+6xy^2-y^3\)

Answer 5

c) \(\left(x^2+x+1\right)\left(x-1\right)\)

\(=x^3-1\)

Answer 6

Tính là sao bạn?