Các câu hỏi tương tự
bài 4 cmr nếu a/b=c/d thì
a. 5a+3b/5a-3b=5c+3d/5c-3d
b.7a^2+3ab/11a^2-8b^2/7c^2+3cd/11c^2-8b^2
Chứng minh nếu \(\frac{a}{b}=\frac{c}{d}\) thì
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\) thì
a,\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b,\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Chứng minh nếu \(\frac{a}{b}=\frac{c}{d}\) thì :
a, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
chứng minh rằng :
\(\frac{a}{b}=\frac{c}{d}\)thì
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}\)= \(\frac{7c^2+3cd}{11c^2-8d^2}\)
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, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Bài 1: biết\(\frac{a}{b}=\frac{c}{d}CMR\)
\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR:
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^3+3ab}{11a^3-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR neu \(\frac{a}{b}=\frac{c}{d}\) thi
a)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)