Question

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Tính :

\(D=\frac{a^3+3^3}{b^3+4^3}\text{ biết }\frac{a+b}{a-3}=\frac{b+4}{b-4}\)

Answer 1

[ミ★Zero ❄ ( Hoàng Nhật )]

Sửa đề \(D=\frac{a^3+3^3}{b^3+4^3}\)biết \(\frac{a+3}{a-3}=\frac{b+4}{b-4}\)

\(\Leftrightarrow\left(a+3\right)\left(b-4\right)=\left(a-3\right)\left(b+4\right)\)

\(\Leftrightarrow ab-4a+3b-12=ab+4a-3b-12\)

\(\Leftrightarrow8a=6b\)

\(\Leftrightarrow\frac{a}{6}=\frac{b}{8}\Leftrightarrow\frac{a}{3}=\frac{b}{4}\)

Đặt \(\frac{a}{3}=\frac{b}{4}=k\)\(\Rightarrow a=3k,b=4k\)

\(\Rightarrow D=\frac{a^3+3^3}{b^3+4^3}=\frac{\left(3k\right)^3+3^3}{\left(4k\right)^3+4^3}\)

\(=\frac{3^3\left(k^3+1\right)}{4^3\left(k^3+1\right)}=\frac{3^3}{4^3}=\frac{27}{64}\)

Answer 2

[Phan Tuấn Anh]

TL: 
8 nhé 

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