à quên
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{5a}{5c}=\frac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số = nhau
\(\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
Chứng minh nếu \(\frac{a}{b}=\frac{c}{d}\) thì :
a, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
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,\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
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) \(\frac{7a^3+3ab}{11a^3-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
chứng minh rằng :
\(\frac{a}{b}=\frac{c}{d}\)thì
a) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}\)= \(\frac{7c^2+3cd}{11c^2-8d^2}\)
\(Cho\)\(\frac{a}{b}=\frac{c}{d}.CMR:\)
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)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
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, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
