Question

Cho a,b,c là 3 số thực khác , thỏa mãn điều kiện: \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

Tính giá trị biểu thực P=\(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

Answer 1

Áp dụng tính chất của dãy tỉ số = nhau ta có:

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

Xét 2 trường hợp:

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

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

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

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

TH2: a + b + c \(\ne\) 0

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

\(\Rightarrow\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}\)\(\Rightarrow\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}\)

\(P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=8\)

 

Answer 2

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

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

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

a+b-c=c

2c=a+b

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

b+c-a=a

2a=b+c

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

c+a-b=b

=>c+a=2b

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

=\(\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)