Từ tỉ lệ thức : \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)(a,b,c,d≠0;a≠±b;c≠±d)ab=cd(a,b,c,d≠0;a≠±b;c≠±d), hãy suy ra các tỉ lệ thức sau:(bằng cách đặt K)
A)\(\dfrac{a+b}{a}\)=\(\dfrac{c+d}{c}\) B)\(\dfrac{a-b}{a}\)=\(\dfrac{c-d}{c}\) C)\(\dfrac{a}{a+b}\)=\(\dfrac{c}{c+d}\) D)\(\dfrac{a}{a-b}\)=\(\dfrac{c}{c-d}\)
Mai mình nộp rồi. Ai làm sớm nhất mình sẽ tick
Xin chân thành cảm ơn
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a=bk;c=dk\)
\(\dfrac{a+b}{a}=\dfrac{bk+b}{bk}=\dfrac{b\left(k+1\right)}{bk}=\dfrac{k+1}{k}\)
\(\dfrac{c+d}{c}=\dfrac{dk+d}{dk}=\dfrac{d\left(k+1\right)}{dk}=\dfrac{k+1}{k}\)
\(\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\rightarrowđpcm\)
\(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\)
\(\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\rightarrowđpcm\)
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\rightarrowđpcm\)
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)
\(\Rightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\rightarrowđpcm\)
