Ta có ; \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}\left(1\right)\)
Mặt khác ; \(\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\left(2\right)\)
Từ : (1) và (2) => \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
Suy ra ; \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
\(Cho\frac{a}{b}=\frac{c}{d}.Tính\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(cho\frac{a}{b}=\frac{c}{d}CMR\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a, \(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
c, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh
a) \(\frac{a.b}{c.d}=\left(\frac{a-b}{c-d}\right)^2\)
b) \(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\)
c) \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
Chứng minh 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 \(\frac{a}{b}=\frac{c}{d}\)thì:
a)\(\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
b)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
c)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho tỉ lệ thức:\(\frac{a}{b}=\frac{c}{d}\left(a;b;c;d\ne0\right)\)
Chứng minh:
a) \(\frac{a+b}{b}=\frac{c+d}{d}\)
b) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
từ \(\frac{a}{b}\)=\(\frac{c}{d}\)chứng minh rằng;
\(\frac{a}{a-b}\)=\(\frac{c}{c-d}\)\(\frac{a+b}{b}=\frac{c+d}{d}\)\(\frac{5a+3b}{5a-3b}=\frac{5c=3d}{5c-3d}\)\(\frac{^{b^2}}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)Chứng minh nếu \(\frac{a}{b}=\frac{c}{d}\) thì
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
