Từ giả thiết ta suy ra ab=c2
Thay số vào ta có : \(\frac{a^2+ab}{b^2+ab}=\frac{a\left(a+b\right)}{b\left(b+a\right)}=\frac{a}{b}\)
=> đcpcm
__cho_mình_nha_chúc_bạn_học _giỏi__
cho \(\frac{a}{c}=\frac{c}{b}\).Chung minh rang: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)
cho \(\frac{a}{c}=\frac{c}{b}\) chung minh rang \(\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\)
Chung minh rang neu a^2=bc thi
a) \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)
b) \(\frac{a^2+c^2}{b^2+a^2}=\frac{c}{b}\)
cho\(b^2=ac\)chung minh rang\(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
Cho \(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\). Chung to rang \(\frac{a}{b}=\frac{a-c}{c-b}\)
chung minh rang:\(\frac{a}{b}=\frac{c}{d}\) thì\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{ab}{c\text{d}}\)
cho \(\frac{a}{b}=\frac{c}{d}\)chung minh
\(\frac{\left(a+b\right)^2}{c+a}=\frac{\left(a^2+b\right)^2}{c+a}\)
cho\(\frac{a}{b}=\frac{b}{c}=\frac{b}{d}\)chung minh rang\(a=b=c\)
Biet \(\frac{a+b}{c-a}=\frac{c+a}{c-a}\) Chung minh rang a2=bc. Dieu nguoc lai co dung khong?
