Cho \(\frac{a}{b}=\frac{c}{d}\) CMR :
A) (a + c ) . ( b - d ) = ( a -c ) . ( b + d )
b) (2a + 3c ) .( 2b - 3d ) = ( 2a - 3c ) . ( 2b + 3d )
1.CHO \(\frac{2A+3C}{2B+3D}=\frac{2X-3C}{2B-3D}CMR:\frac{A}{B}=\frac{C}{D}\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C+2A-3C}{2B+3D+2B-3D}=\frac{4A}{4B}=\frac{A}{B}\left(1\right)\)\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C-2A+3C}{2B+3D-2B+3D}=\frac{6C}{6D}=\frac{C}{D}\left(2\right)\)
Từ (1) và (2) suy ra : \(\frac{A}{B}=\frac{C}{D}\)
Giải :
Từ đảng thức : \(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)
\(\Rightarrow\left(2a+3c\right).\left(2b-3d\right)=\left(2b+3d\right).\left(2a-3c\right)\)
\(\Rightarrow4ab-6ad+6bc-9cd=4ab-6bc+6ad-9cd\)
\(\Rightarrow\left(4ab-6ad+6bc-9cd\right)-\left(4ab-6bc+6ad-9cd\right)=0\)
\(\Rightarrow4ab-6ad+6bc-9cd-4ab+6bc-6ad+9cd=0\)
\(\Rightarrow\left(4ab-4ab\right)-\left(6ad+6ad\right)+\left(6bc+6bc\right)-\left(9cd-9cd\right)=0\)
\(\Rightarrow-12ad+12bc=0\)
\(\Rightarrow12bc=12ad\)
\(\Rightarrow bc=ad\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)
Cho a/b=c/d. CMR : 2a-3c/2b-3d=2a+3c/2a+3d
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Khi đó:
\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)
\(\frac{2a+3c}{2a+3d}=\frac{2bk+3dk}{2a+3d}=\frac{k\left(2a+3d\right)}{2a+3d}=k\)
Vậy \(\frac{2a-3c}{2b-3d}=\frac{2a+3c}{2a+3d}=k\)
Ta có đpcm
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 a/b=c/d c/m 2a+3c/2a-3c=2b+3d/2b-3d
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{2a}{2b}=\dfrac{3c}{3d}=\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
\(\Rightarrow\dfrac{2a+3c}{2a-3c}=\dfrac{2b+3d}{2b-3d}\)
\(\Rightarrow dpcm\)
cho \(\frac{a}{b}\)=\(\frac{c}{d}\)CMR \(\frac{2a-3c}{2b-3d}\)=\(\frac{2a+3c}{2a+3d}\)
Vì \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=kd\)
\(\Rightarrow\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\)(1)
\(\Rightarrow\frac{2a+3c}{2b+3d}=\frac{2bk+3dk}{2b+3d}=\frac{k\left(2b+3d\right)}{2b+3d}=k\)(2)
\(\RightarrowĐPCM\)
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/\(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3b}\left(2b+3d\ne0;2b-3d\ne0\right)\)
Tim\(\frac{a}{b}=\frac{c}{d}CMR\frac{a}{b}=\frac{c}{d}=\frac{2a-3c}{2b-3d}\)
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)
Cho a/b=c/d Với b/d khác +-3/2 . Chứng minh rằng:
a)2a+3c/2b+3d=2a-3c/2b-3d.
b)a^2+c^2/b^2+d^2=ac/bd
Cho các phân số a/b;c/d. Biết ab=cd, chứng minh rằng \(\frac{2a-3c}{2b-3d}=\frac{2a+3c}{2a+3d}\)