Question

BT6: Thu gọn về hàng đẳng thức

\(1,\left(2x+1\right)^2+2\left(2x+1\right)+1\)

\(2,\left(3x-2y\right)^2+4\left(3x-2y\right)+4\)

Answer 1

\(1,\left(2x+1\right)^2+2\left(2x+1\right)+1\\ =\left(2x+1\right)^2+2.\left(2x+1\right).1+1^2\\ =\left[\left(2x+1\right)+1\right]^2\\ b,\left(3x-2y\right)^2+4\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2+2.\left(3x-2y\right).2+2^2\\ =\left[\left(3x-2y\right)+2\right]^2\)

Answer 2

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

\(=\left(2x+1\right)^2+2\left(2x+1\right)\cdot1+1^2\)

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

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

2) \(\left(3x+2y\right)^2+4\left(3x+2y\right)+4\)

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

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

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