Question

Cho \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}\). Tính \(\frac{4a+2b-c}{a-b-c}\)

Answer 1

Đặt \(\frac{a}{2}\)\(=\)\(\frac{b}{5}\)\(=\)\(\frac{c}{7}\)\(=\)K

=> a=2K

     b=5k

     c=7k

=> \(\frac{4a+2b-c}{a-b-c}\)= \(\frac{8k+10k-7k}{2k-5k-7k}\)= \(\frac{k.\left(8+10-7\right)}{k.\left(2-5-7\right)}\)= \(\frac{8+10-7}{2-5-7}\)=  \(\frac{-11}{10}\)

Answer 2

Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\)

\(\Rightarrow a=2k\)

\(b=5k\)

\(c=7k\)

\(\Rightarrow\frac{4a+2b-c}{a-b-c}\)

\(=\frac{4\left(2k\right)+2\left(5k\right)-7k}{2k-5k-7k}\)

\(=\frac{\left(8+10-7\right)k}{\left(2-5-7\right)k}\)

\(=-\frac{11}{10}\)

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