cho \(\frac{a}{b}\) = \(\frac{c}{d}\) chứng minh rằng :\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR: nếu \(\frac{a}{b}=\frac{c}{d}thì\left(a\right)\frac{5a+3b}{5a-3b}-\frac{5c+3d}{5c-3d}\)
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) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
CMR: Nếu a/b=c/d thì 7a2+3ab/11a2-8b2=7c2+3cd/11c2-8d2
CMR: Nếu a2=bc thì a+b/a-b=c+a/c-a
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Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) CMR:
a) \(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)
b) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
các bn ơi giúp mình với mai mình hok rồi
Cho \(\frac{a}{b}\) = \(\frac{c}{d}\). Chứng minh:
a) \(\frac{11a+3b}{11c+3d}\) = \(\frac{3a-11b}{3c-11d}\)
b) \(\frac{a^2+c^2}{^{ }b^2+d^2}\) = \(\frac{ac}{bd}\)
c) \(\frac{4a^2+5b^4}{4c^4+5d^4}\) = \(\frac{a^2b^2}{c^2d^2}\)
giúp mình với các bn ơi
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bài 1; cho a/b =c/d
a) \(\frac{a-b}{b}=\frac{c-d}{d}\)
b)\(\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)
c)\(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)
d)\(\frac{4a^4+5b^4}{4a^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
Rút gọn: \(\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
0\(0\le a\le b\le c\le1\)chứng minh rằng \(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}< 2\)HELP ME!!!!!!!!!!!!!!!!