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Ta có:\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a}{b}\cdot\frac{b}{c}=\frac{a}{c}\)
Mà\(\frac{a}{b}=\frac{b}{c}=\frac{2015b}{2015c}=\frac{a+2015b}{b+2015c}\)
Nên suy ra\(\frac{a}{c}=\frac{a^2}{b^2}=\left(\frac{a+2015b}{b+2015c}\right)^2=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\)
Vậy\(\frac{a}{c}=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\left(đpcm\right)\)
vì b2 = a.c nên \(\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a}{b}=\frac{2015.b}{2015.c}=\frac{a+2015.b}{b+2015.c}\)
\(\Rightarrow\left(\frac{a+2015.b}{b+2015.c}\right)^2=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}=\frac{a^2}{a.c}=\frac{a}{c}\)
Từ : \(b^2=a\Rightarrow\frac{a}{b}=\frac{b}{c}\)
Hay \(\frac{a}{b}=\frac{b}{c}=\frac{2016b}{2016c}\)
Áp dụng tích chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{2016b}{2016c}=\frac{a+2016b}{a+2016c}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{a+2016b}{b+2016c}\)
\(\Rightarrow\frac{a}{b}.\frac{b}{c}=\left(\frac{a+2016b}{a+2016c}\right)^2\)
Hay \(\frac{a.b}{b.c}=\frac{\left(a+2016b\right)^2}{\left(b+2016c\right)^2}\Rightarrow\frac{a}{c}=\frac{\left(a+2016b\right)^2}{\left(b+2016c\right)^2}\)(ĐPCM)
mk nha