Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
=>\(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
Chứng minh rằng nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì
a, \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5a-3d}\)
b, \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Bài 4: Chứng minh rằng:
Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì:
a) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b)\(\dfrac{7a^2+3ab}{11b^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Các bạn giúp mềnh nha, mai mềnh đi học òy, rồi mềnh tick cho
Thank các bạn
a, CMR : nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\) ( gt các tỉ số đều có nghĩa )
b, tìm x,biết: \(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}+\dfrac{x-3}{2015}=\dfrac{x-4}{2014}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng : \(\frac{5a+3b}{5a-3b}\) \(=\frac{5c+3d}{5c-3d}\)
Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) Chứng minh rằng: \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng:
a) \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
b) \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
Help meeee!!!
CMR: Nếu \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) thì \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh \(\dfrac{a}{b} = \dfrac{c}{d}\) nếu biết
a, \(\dfrac {4a-3b}{4c-3d} = \dfrac {4a+3b}{4c+3d}\)
b, \(\dfrac {2a-3b}{2a+3b} = \dfrac {2c-3d}{2c+3d}\)
Cho tỉ lệ thức \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng:
a)\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
b)\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . CMR :
a, \(\dfrac{5a+3b}{7a-2b}=\dfrac{5c+3d}{7c-2d}\)
b, \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
c, \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
( giả thiết các tỉ số trên đều có nghĩa )
