đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
=>\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k.\left(2b-3d\right)}{2b-3d}=k\)
suy ra:\(\frac{a}{b}=\frac{c}{d}=\frac{2a-3c}{2b-3d}\)( vì cùng = k)
1.CHO \(\frac{2A+3C}{2B+3D}=\frac{2X-3C}{2B-3D}CMR:\frac{A}{B}=\frac{C}{D}\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR :
a/ \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
b/\(\frac{2a+3c}{2d+3d}=\frac{2a-3c}{2b-3d}\)
c/\(\frac{a^2+c^2}{b^2+d^2}=\frac{ac^2}{bd}\)
cho tỉ lệ thức ;\(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng ;
a/\(\frac{a+b}{b}=\frac{c+d}{d}\)
b/\(\frac{a}{a+b}=\frac{c}{c+d}\left(a+b#0;c+d#0\right)\)
c/\(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3b}\left(2b+3d\ne0;2b-3d\ne0\right)\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\).Hãy suy ra :
a/ \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
b/\(\frac{2a+3c}{2d+3d}=\frac{2a-3c}{2b-3d}\)
c/ \(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) Chung minh cac ti le thuc sau :
a) \(\frac{2a+c}{2b+d}=\frac{2a-c}{2b-d}\)
b) \(\frac{2a+3c}{a}=\frac{2b+3d}{b}\)
Bài 2: Từ \(\frac{a}{b}=\frac{c}{d}\)hãy chứng minh: \(\frac{a}{b}=\frac{c}{d}=\frac{2a-3c}{2b-3d}\)
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\),chứng minh rằng \(\frac{2a-3c}{2b-3d}\)=\(\frac{2a+3c}{2a+3d}\)
\(Cho\) \(\frac{a}{b}=\frac{c}{d}\)
\(CMR:\)\(a,\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)
\(b,\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\)
Cho các phân số a/b;c/d. Biết ab=cd, chứng minh rằng \(\frac{2a-3c}{2b-3d}=\frac{2a+3c}{2a+3d}\)
