Question

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

Tính:M=\(\frac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\)

Answer 1

Ta có

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

\(=2\)

Từ \(\frac{2b+c-a}{a}=2\Rightarrow2a=2b+c-a\Rightarrow3a-2b=c\)và \(3a-c=2b\)

Tương tự có \(3b-2c=a;3b-a=2c\) và \(3c-2a=b;3c-b=2a\)

Thay vào biểu thức M ta có

\(M=\frac{a\cdot b\cdot c}{2\cdot b\cdot2\cdot a\cdot2\cdot c}=\frac{1}{8}\)

Answer 2

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