Question

chứn minh rằng

a) (a+b0^3+(a-b)^3=2a(a^2+3b^2)

b) (a+b)^3-(a-b)^3=2b(b^2+3a^2)

Answer 1

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

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3-6ab^2=0\)

\(\Leftrightarrow0=0\) ( đpcm) .

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

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2a^3-6ab^2=0\)

\(\Leftrightarrow0=0\) ( luôn đúng )

Vậy đẳng thức được chứng minh.

Answer 2

Làm cách khác với "thị nở" :v.

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

\(=\left[\left(a+b\right)+\left(a-b\right)\right]\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]=2a\left(a^2+3b^2\right)\)

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

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

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

\(=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]=2b\left(b^2+3a^2\right)\)

\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)=2b\left(b^1+3a^2\right)\)\(=2b^2\left(b^2+3a^2\right)=2b^2\left(b^2+3a^2\right)\)

Answer 3

Chỗ 2b²(b¹+3a²) sửa b¹ thành b² nha. Còn cái nữa là do xuống hàng nên hơi khó nhìn :v