Theo đề ta có: \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)
\(\frac{b^2-a^2}{a^2+c^2}=\frac{b^2-a^2}{a^2+ab}=\frac{\left(b-a\right)\left(a+b\right)}{a\left(a+b\right)}=\frac{b-a}{a}\left(đpmc\right)\)
Các câu hỏi tương tự
Cho a,b,c>0.Chứng minh rằng:
\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
Cho a,b,c>0.Chứng minh rằng:
\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
Cho \(\frac{a}{c}=\frac{c}{b}\)chứng minh rằng:
a)\(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)
b)\(\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\)
Cho \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) chứng minh rằng \(\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\)
Cho a, b, c > 0. Chứng minh rằng: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{9\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\)
1. Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
2. Cho \(\frac{a}{b}=\frac{b}{c}\) chứng minh rằng \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
Cho \(\frac{a}{b}=\frac{b}{c}\).Chứng minh rằng:\(\frac{a}{c}=\frac{a^2+c^2}{b^2+c^2}\)
Cho \(\frac{a}{c}=\frac{c}{b}\)chứng minh rằng:\(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)
cho \(\frac{a}{c}=\frac{c}{b}\). Chứng minh rằng : \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)