Question

CMR từ hệ thức \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) ta có hệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)

Answer 1

Ta có: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Leftrightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}.\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{a+a+b-b}{c+c+d-d}=\dfrac{2a}{2c}=\dfrac{a}{c}_{\left(1\right)}.\)

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{a-a+b+b}{c-c+d+d}=\dfrac{2b}{2d}=\dfrac{b}{d}_{\left(2\right)}.\)

Từ \(_{\left(1\right)+\left(2\right)}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\) (t/c tỉ lệ thức).

\(\Rightarrowđpcm.\)

Answer 2

a=b*k

c=d*k

thì b*k+b/b*k-b=b*(k+1)/b*(k-1)=k+1/k-1

thì d*k+d/d*k-d=d*(k+1)/d*(k-1)=k+1/k-1

nen suy ra a+b/a-b=c+d/c-d