(x+3)^2+2.(x+3).(x-3)+(x-3)^2

$(x+3)^2+2.(x+3).(x-3)+(x-3)^2$

$= [(x+3) + (x-3)]^2$

$= (2x)^2$

$= 4x^2$.

Đáp án:

Giải thích các bước giải: $\begin{array}{l}{(x + 3)^2} + 2.(x + 3).(x - 3) + {(x - 3)^2}\\ = {\left( {x + 3} \right)^2} + 2\left( {x + 3} \right)\left( {x - 3} \right) + \left( {x - 3} \right)\left( {x + 3} \right)\\ = \left( {x + 3} \right)\left( {x + 3 + 2\left( {x - 3} \right) + x - 3} \right)\\ = \left( {x + 3} \right)\left( {x + 3 + 2x - 6 + x - 3} \right)\\ = \left( {x + 3} \right)\left( {4x - 6} \right)\end{array}$