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Lớp 7 gì mà dễ ẹc :))
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Rightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a-5b=-3c\)
\(\Leftrightarrow a-4a=-3c\)
\(\Leftrightarrow-3a=-3c\)
\(\Rightarrow a=c\)
Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)
Ta có: \(\frac{2a+b+c}{a}=\frac{a+2b+c}{b}=\frac{a+b+2c}{c}\)
\(\Rightarrow\frac{2a+b+c}{a}-1=\frac{a+2b+c}{b}-1=\frac{a+b+2c}{c}-1\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Mà \(a,b,c\ne0\)
=> a = b= c
\(A=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}\)
\(=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}\)
\(=2+2+2=6\)
\(\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Rightarrow\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{\left(2a-b\right)+\left(b-c+a\right)}{\left(a+b\right)+\left(2a-b\right)}=\frac{3a-c}{3a}=\frac{2}{3}\)
\(\Rightarrow2\times3a=3\times\left(3a-c\right)\)
\(\Rightarrow6a=9a-3c\)
\(\Rightarrow6a-9a=-3c\)
\(\Rightarrow-3a=-3c\)
\(\Rightarrow\frac{-3a}{-3}=\frac{-3c}{-3}\)
\(\Rightarrow a=c\)
\(\Rightarrow\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(5b+4a\right)^5}{\left(5b+4a\right)^2\left(a+3a\right)^3}=\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Rightarrow3\times\left(2a-b\right)=2\left(a+b\right)\)
\(\Rightarrow6a-3b=2a+2b\)
\(\Rightarrow6a-2a=3b+2b\)
\(\Rightarrow4a=5b\)
\(\Rightarrow b=\frac{4a}{5}\)
\(\Rightarrow\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}=\left(\frac{5\times\frac{4a}{5}+4a}{4a}\right)^3=\left(\frac{4a+4a}{4a}\right)^3\)
\(\Rightarrow\left(\frac{8a}{4a}\right)^3=2^3=8\)