Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\left(k\in R\right)\)
\(\Rightarrow\hept{\begin{cases}a=2k\\b=5k\\c=7k\end{cases}}\)
Thay vào A ta được \(A=\frac{2k-5k+7k}{2k+2\cdot5k-7k}=\frac{4k}{5k}=\frac{4}{5}\)
Vậy A=\(\frac{4}{5}\)
Đặt a/2 = b/5 = c/7 = k => a = 2k
b = 5k
c = 7k
=> a - b + c / a + 2b - c = 2k - 5k + 7k / 2k + 2 * 5k - 7k = 4k / 5k = 4/5
Vậy A = 4/5
Ta có \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=\frac{a-b+c}{2-5+7}=\frac{a-b+c}{4}\)
x-y+z=2a
ta lại có \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=\frac{a+2b-c}{2+10-7}=\frac{a+2b-c}{5}\)
\(a+2b-c=\frac{5a}{2}\)
do đó A =\(\frac{a-b+c}{a+2b-c}=\frac{2a}{\frac{5a}{2}}=\frac{4}{5}\)
Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\)=>a=2k,b=5k,c=7k
Thay vào ta có: A=\(\frac{2k-5k+7k}{2k+2.5k-7k}\)=\(\frac{\left(2-5+7\right)k}{\left(2+10-7\right)k}\)=\(\frac{4k}{5k}\)=\(\frac{4}{5}\)
Từ \(\frac{a}{2}=\frac{4}{5}=>a=\frac{8}{5}\)
\(\frac{b}{5}=\frac{4}{5}=>b=4\)
\(\frac{c}{7}=\frac{4}{5}=>c=\frac{28}{5}\)
