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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\(\frac{a}{c}=\frac{\left(a+2015b\right)^2}{\left(b+2015c\right)^2}\left(đpcm\right)\)
Ta có b^2=ac =>a/b=c/d. Đặt a/b=c/d=k(khác 0) =>a=bk;b=ck =>a/c=c.k^2/c=k^2 (1) (a+2015b)^2/(b+2015c)^2=(bk+2015b/ck+2015c)^2=(b(k+2015)/(c(k+2015))^2=(b/c)^2=(ck/c)^2=k^2 (2) Từ (1) và (2) => a/c=(a+2015b/b+2015c)^2 => (đpcm)
Ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\iff\)\(\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)
\(\iff\) \(ac+bc=ab+ac=bc+ba\)
+)\(ac+bc=ab+ac\)
\(\implies\)\(bc=ab\)
\(\implies\) \(c=a\left(1\right)\)
+)\(ab+ac=bc+ba\)
\(\implies\) \(ac=bc\)
\(\implies\) \(a=b\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)
\(\implies\) \(a=b=c\)
\(\implies\) \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{aa+bb+cc}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Vậy \(M=1\)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
=> \(\frac{a}{b}=\frac{2013b}{2013c}=\frac{a+2013b}{b+2013c}\)
=> \(\frac{a}{b}.\frac{b}{c}=\frac{a+2013b}{b+2013c}.\frac{a+2013b}{b+2013c}\Rightarrow\frac{a}{c}=\left(\frac{a+2013b}{b+2013c}\right)^2\)
Ta có: \(b^2=a.c\)
Suy ra: \(b.b=a.c\)
Suy ra: \(\frac{a}{b}=\frac{b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{2013b}{2013c}=\frac{a+2013b}{b+2013c}\)
Khi đó: \(\frac{\left(a+2013b\right)^2}{\left(b+2013c\right)^2}=\left(\frac{a+2013b}{b+2013c}\right)^2=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}\)
<=> \(\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)
<=> \(\frac{9a}{a+b}=\frac{9b}{b+c}\)
<=> \(\frac{a}{a+b}=\frac{b}{b+c}\)
=> a(b + c) = b(a + b)
<=> ab + ac = ba + b2
=> ac = b2 (đpcm)
