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Câu hỏi của Tăng Thiện Đạt - Toán lớp 8 - Học toán với OnlineMath

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

<=> \(\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1\)

<=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

<=> a + b + c = 0 hoặc a = b = c.

Th1: a + b + c = 0 

=> a + b = - c ; a + c = -b ; b + c = -a.

Thế vào P :

\(P=\left(1+\frac{a}{b}\right)\cdot\left(1+\frac{b}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)

\(=\left(\frac{a+b}{b}\right)\cdot\left(\frac{b+c}{c}\right)\cdot\left(\frac{c+a}{a}\right)\)

\(=-\frac{c}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)

TH2: a = b = c. THế vào P 

\(P=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)

Vậy: P = -1 nếu a + b + c = 0 

hoặc P = 8 nếu a = b = c.

\(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)

Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)\(\Rightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1=\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

\(\Rightarrow P=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{\left(-a\right).\left(-b\right).\left(-c\right)}{abc}=-1\)

TH2: Nếu \(a+b+c\ne0\)\(\Rightarrow a=b=c\)

\(\Rightarrow\hept{\begin{cases}a+b=2b\\b+c=2c\\c+a=2a\end{cases}}\)\(\Rightarrow P=\frac{2b}{b}.\frac{2c}{c}.\frac{2a}{a}=2.2.2=8\)

Vậy \(P=-1\)hoặc \(P=8\)

\(\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)

= \(\left(1+\frac{\left(b+c\right)^2-2bc-a^2}{2bc}\right).\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{\left(b+c\right)^2-2bc-\left(b-c\right)^2}{a+b+c}\)

= \(\left(1+\frac{\left(b+c-a\right)\left(b+c+a\right)-2bc}{2bc}\right).\frac{a+b+c}{b+c-a}.\frac{\left(b+c-b+c\right)\left(b+c+b-c\right)-2bc}{a+b+c}\)

= \(\left(1+\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}-1\right).\frac{a+b+c}{b+c-a}.\frac{4bc-2bc}{a+b+c}\)

= \(\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}.\frac{2bc}{b+c-a}\)

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

= \(b+c+a\)

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

\(\Leftrightarrow a+b=2c=b+c=2a=a+c=2b\Rightarrow a=b=c\)

\(M=\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)=2^3=8\)

Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath