Question

Cho \(\dfrac{a+b}{a-3}=\dfrac{b+4}{b-4}\)Timhs giá trị biểu thức D=\(\dfrac{a^3+3^3}{b^3+4^3}\)

Answer 1

Sửa đề: \(\dfrac{a+3}{a-3}=\dfrac{b+4}{b-4}\)

=>(a+3)(b-4)=(a-3)(b+4)

=>ab-4a+3b-12=ab+4a-3b-12

=>-4a+3b=4a-3b

=>-8a=-6b

=>\(4a=3b\)

=>\(\dfrac{a}{3}=\dfrac{b}{4}=k\)

=>a=3k; b=4k

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

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