Question

BÀI 1:    Cho \(ac=b^2;bd=c^2\)

CMR: \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)

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

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

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Answer 1

BÀI 2: Áp dụng tc của dãy tỉ số bằng nhau, ta có:

\(\frac{2a+b+c}{a}=\frac{a+2b+c}{b}=\frac{a+b+2c}{c}=\frac{4a+4b+4c}{a+b+c}=4\) 

\(\Rightarrow2+\frac{b+c}{a}=2+\frac{a+c}{b}=2+\frac{a+b}{c}=4\)

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

Vậy N = 6

Answer 2

BÀI 1: Theo đề bài, ta có:

\(ac+c^2=b^2+bd\Rightarrow c\left(a+c\right)=b\left(b+d\right)\Rightarrow c\left(a+c\right)+bc=b\left(b+d\right)+bc\)\(\Rightarrow c\left(a+b+c\right)=b\left(b+c+d\right)\)\(\Rightarrow\frac{a+b+c}{b+c+d}=\frac{b}{c}\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\left(\frac{b}{c}\right)^3=\frac{b^2b}{c^2c}=\frac{acb}{bdc}=\frac{a}{d}\).