7 + 7 + 7 + 7 + 7 = 7 x 5 = 35

8 + 8 + 8 + 8 + 8 = 8 x 5 = 40

a) Xét vế trái: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+3a^2bc+3abc^2+c^3-a^3-b^3-c^3\)

\(=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3\)

\(=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

tốn S quá

Ta có:|x + 2| - x = 2

   <=> |x + 2|  = 2 + x 

    => x + 2 = 2 + x