Question

Tính P = \(\frac{a+2b}{3a}+\frac{b+2a}{3b}\)    khi \(a^2-3ab+2b^2=0\)

 

Answer 1

\(a^2-3ab+2b^2=0\)

\(\Leftrightarrow a^2-2ab-ab+2b^2=0\)

\(\Leftrightarrow a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=2b\\a=b\end{cases}}\)

+ ) TH1 :  

\(a=2b\)

\(P=\frac{a+2b}{3a}+\frac{b+2a}{3b}\)

\(P=\frac{2b+2b}{6b}+\frac{b+4b}{3b}\)

\(P=\frac{4b}{6b}+\frac{5b}{3b}\)

\(P=\frac{4}{6}+\frac{5}{3}=\frac{7}{3}\)

+ ) TH 2  \(a=b\)

\(P=\frac{a+2b}{3a}+\frac{b+2a}{3b}\)

\(P=\frac{3a}{3a}+\frac{3b}{3b}=1+1=2\)

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