Question

[Lê Hoàng Nam]

Câu 1: Cho a+b-c/c=b+c-a/a=c+a-b

Tính M=(1+b/a)*(1+a/c)*(1+c/b)

Answer 1

[Xyz OLM]

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

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

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

Nếu a + b + c = 0

=> a + b = - c ;

a + c = - b

b + c = - a 

Khi đó M = \(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-\left(abc\right)}{abc}=-1\)

Nếu a +b + c \(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)

Khi đó M = a + b/a . a + c/c . b + c/b = 2a/a . 2c/c . 2b/b =  2.2.2 = 8 

Vậy M = 8 hoặc M = - 1

Answer 2

[Trần_Hiền_Mai]

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

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

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

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

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

TH1) (trường hợp 1) \(a+b+c\ne0\)\(\Rightarrow\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\frac{a+b}{c}-1=\frac{a+c}{b}-1=\frac{b+c}{a}-1=1\)

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

Ta có: \(M=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)\(=\left(\frac{a}{a}+\frac{b}{a}\right)\left(\frac{c}{c}+\frac{a}{c}\right)\left(\frac{b}{b}+\frac{c}{b}\right)\)

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

               \(=\frac{a+b}{c}.\frac{a+c}{b}.\frac{b+c}{a}=2.2.2=8\)

TH2) (trường hợp 2) \(a+b+c=0\)

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

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

    Vậy, M= 8 hoặc M=-1

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