Question

Bài 1: Tính

a) (2x + 1 + 4y)²

b) (2x + 1)² + (3x - 1)² + 2 . (2x + 1) . (3x - 1)

Answer 1

`a)(2x+1+4y)^2`

`=(2x)^2+1^2+(4y)^2+2.2x.1+2.2x.4y+2.1.4y`

`=4x^2+1+16y^2+4x+16xy+8y`

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`b)(2x+1)^2+(3x-1)^2+2(2x+1)(3x-1)`

`=4x^2+4x+1+9x^2-6x+1+2(6x^2-2x+3x-1)`

`=13x^2-2x+2+12x^2-4x+6x-2`

`=25x^2`

Answer 2

a, \(=\left[\left(2x+1\right)+4y\right]^2=\left(2x+1\right)^2+8y\left(2x+1\right)+16y^2\)

\(=4x^2+4x+1+16xy+8y+16y^2\)

b, \(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)