Question

tìm a,b,c biết 

a/b=2/3 , a/c=1/2 và a^3+b^3+c^3=99

Answer 1

\(\dfrac{a}{b}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3};\dfrac{a}{c}=\dfrac{1}{2}\Rightarrow\dfrac{a}{1}=\dfrac{c}{2}\\ \Rightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}\)

Áp dụng tcdtsnb:

\(\dfrac{a^3}{8}=\dfrac{b^3}{27}=\dfrac{c^3}{64}=\dfrac{a^3+b^3+c^3}{8+27+64}=\dfrac{99}{99}=1\\ \Rightarrow\left\{{}\begin{matrix}a^3=8\\b^3=27\\c^3=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)