Question

Cho a, b, c là các số thỏa mãn: a+b=c. Chứng minh \(\frac{a^3-b^3}{a^3+c^3}\)= \(\frac{a-b}{a+c}\)

Answer 1

Akai Haruma em có cách khác cô nè:)

\(\frac{a^3-b^3}{a^3+c^3}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\) (1)

Cần chứng minh \(a^2+ab+b^2=a^2-ac+c^2\Leftrightarrow ab+b^2=c^2-ac\)

\(\Leftrightarrow b\left(a+b\right)=c\left(c-a\right)\Leftrightarrow b\left(a+b\right)=\left(a+b\right)\left(a+b-a\right)\)

\(\Leftrightarrow b\left(a+b\right)=b\left(a+b\right)\) (đúng)

Do vậy \(\frac{a^3-b^3}{a^3+c^3}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}=\frac{a-b}{a+c}^{\left(đpcm\right)}\)

Answer 2

Lời giải:

Ta có:
\(a^3-b^3=(a-b)(a^2+ab+b^2)\)

\(a^3+c^3=(a+c)(a^2-ac+c^2)=(a+c)[a^2-a(a+b)+(a+b)^2]\) (thay $c=a+b$)

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

Do đó:

\(\frac{a^3-b^3}{a^3+c^3}=\frac{(a-b)(a^2+ab+b^2)}{(a+c)(a^2+ab+b^2)}=\frac{a-b}{a+c}\)

Ta có đpcm.