Ta có :
\(b^2=ac\)\(\Rightarrow\)\(\frac{a}{b}=\frac{b}{c}\)\(\Rightarrow\)\(\frac{a}{b}=\frac{2017b}{2017c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\)\(\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)\(\left(1\right)\)
Lại có :
\(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{ab}{bc}=\frac{a}{c}\)\(\left(2\right)\)
Từ (1) và (2) suy ra :
\(\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)
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Ta có: \(b^2=ac\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a}{b}=\frac{2017b}{2017c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\left(1\right)\)
Ta lại có:
\(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{ab}{bc}=\frac{a}{c}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}=\frac{a}{c}\)
