Question

CMR Nếu a = b + c thì \(\dfrac{a^3+b^3}{a^3+c^3}=\dfrac{a+b}{a+c}\)

Answer 1

Xét \(a+b=0\) thì ta có ĐPCM

Xét \(b=c\)

\(\Rightarrow a=2c\)

Ta chứng minh:

\(\dfrac{8c^3+c^3}{8c^3+c}=\dfrac{2c+c}{2c+c}\)

\(\Leftrightarrow1=1\) đúng

Xét \(\left\{{}\begin{matrix}a+b\ne0\\b\ne c\end{matrix}\right.\)

Ta chứng minh:

\(\dfrac{a^3+b^3}{a^3+c^3}=\dfrac{a+b}{a+c}\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}=\dfrac{a+b}{a+c}\)

\(\Leftrightarrow\dfrac{a^2-ab+b^2}{a^2-ac+c^2}=1\)

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

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

\(\Leftrightarrow a=c+b\) đúng

Vậy ta có ĐPCM