Đặt \(\frac{a}{b}\) = \(\frac{c}{d}\)=k
=> a = b.k ; c= d.k
=> \(\frac{5a+3b}{5a-3b}\)= \(\frac{5bk+3b}{5bk-3b}\)= \(\frac{b\left(5k+3\right)}{b\left(5k-3\right)}\)= \(\frac{5k+3}{5k-3}\)(1)
\(\frac{5c+3d}{5c-3d}\)= \(\frac{5dk+3d}{5dk-3d}\)= \(\frac{d\left(5k+3\right)}{d\left(5k-3\right)}\)=\(\frac{5k+3}{5k-3}\)(2)
Từ 1 và 2 suy ra điều cần chứng minh
Chứng minh rằng: Nếu \(\frac{a}{c}=\frac{b}{d}\)thì \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
cho \(\frac{a}{b}=\frac{c}{d}\left(c\ne+-\frac{3}{5}d\right)\)
chứng minh \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
Chứng minh rằng nếu
\(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
Cho \(\frac{a}{b}=\frac{c}{d}\ne+-1\)và \(c\ne0\). Chứng minh rằng \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
chứng minh rằng \(\frac{a}{b}=\frac{c}{d}\)thì
\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(Cho\frac{a}{b}=\frac{c}{d}.Tính\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(cho\frac{a}{b}=\frac{c}{d}CMR\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\)
thì: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng : \(\frac{5a+3b}{5a-3b}\) \(=\frac{5c+3d}{5c-3d}\)
Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) Chứng minh rằng: \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
