\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(1\right)\)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)
Từ (1) và (2) => đpcm
Chứng minh rằng nếu
\(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\)
thì: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
chứng minh rằng \(\frac{a}{b}=\frac{c}{d}\)thì
\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
Chứng minh rằng nếu\(\frac{a}{b}\)=\(\frac{c}{d}\) thì:\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\) thì
a,\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b,\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR nếu a/b=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 \(\frac{a}{b}=\frac{c}{d}\)
a, \(\frac{5a+9c}{5b+9d}=\frac{2a}{2b}\)
b, \(\frac{5a+3b}{3a-3b}=\frac{5c+3d}{5c-3d}\)
giúp mik với ạ
