Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=b*k; c=d*k
=>\(\dfrac{a+b}{c+d}=\dfrac{b\cdot k+b}{d\cdot k+d}=\dfrac{b\cdot\left(k+1\right)}{d\cdot\left(k+1\right)}=\dfrac{b}{d}\)(1)
=>\(\dfrac{a+2b}{c+2d}=\dfrac{b\cdot k+2b}{d\cdot k+2d}=\dfrac{b\cdot\left(k+2\right)}{d\cdot\left(k+2\right)}=\dfrac{b}{d}\) (2)
Từ (1)và(2) => \(\dfrac{a+b}{c+d}=\dfrac{a+2b}{c+2d}\)
Bài 1: Cho tỉ lệ thức: \(\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}\)( \(b,d\ne0\)) CMR: \(\dfrac{a}{b}=\dfrac{c}{d}\)
Bài 2: Cho \(yz:zx=1:2\) Hãy tính \(\dfrac{x}{y^2}:\dfrac{y}{2^x}\)
Chứng tỏ rằng tử đẳng thức \(\left(a-2c\right)\left(b+2d\right)=\left(b-2d\right)\left(a+2c\right)\) ta suy ra tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\left(a,b,c,d\ne0\right)\)
Cho a/b=c/d. Hay chung to:
\(\dfrac{2d-3c}{d}=\dfrac{2b-3a}{b}\)
Cho tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng các tỉ lệ thức:
(Giả thiết rằng các tỉ lệ thức cần chứng minh đều có nghĩa)
a) \(\dfrac{a+2b}{2a-b}\)=\(\dfrac{c+2d}{2c-d}\) , b) (a+3c).(b-d)=(a-c).(b+3d)
Chứng minh rằng: Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
Chứng minh rằng : Nếu \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) thì
a.\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\) b.\(\dfrac{a}{b}\)=\(\dfrac{a+c}{b+c}\) c.\(\dfrac{a}{c}\)=\(\dfrac{a-b}{c-d}\) d.\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)
cmr:\(\dfrac{a}{b}=\dfrac{c}{d}\)
nếu \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) .Cmr:
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\) \(\dfrac{a-c}{a+c}=\dfrac{b-d}{b+d}\)
\(\dfrac{a}{a+c}=\dfrac{b}{b+d}\) \(\dfrac{a}{a-c}=\dfrac{b}{b-d}\)
Bài 1 Cho \(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\left(b\ne0\right)\) CMR \(c=0\)
Bài 2 Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}CMR\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
