Bài này bạn chỉ cần đặt k rồi thế k vào là làm được à, dễ lắm
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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 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\(\frac{a}{b}=\frac{c}{d}CMR\)
\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR neu \(\frac{a}{b}=\frac{c}{d}\) thi
a)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR:
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^3+3ab}{11a^3-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
\(Cho\)\(\frac{a}{b}=\frac{c}{d}.CMR:\)
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)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)