Question

cho \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\)

Hãy chứng minh :\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) (các tỉ số đều có nghĩa)

Các bạn jup mình vơi càng nghĩ được nhiều cách càng tốt !!!

Answer 1

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\left(1\right)\\ \Rightarrow a=bk;c=dk\)

Ta có:

\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\left(b+d\ne0\right)\\ \dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\left(3\right)\left(b-d\ne0\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)

Vậy \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\left(dpcm\right)\)

Answer 2

Đặt :\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\) Ta có :\(\dfrac{a+b}{c+d}=\dfrac{bk+b}{dk+d}=\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}=\dfrac{b}{d}\left(1\right)\) \(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b.\left(k-1\right)}{d.\left(k-1\right)}=\dfrac{b}{d}\left(2\right)\) Từ (1) và (2) suy ra : \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) (đpcm)