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a) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) và \(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\\\dfrac{c-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\end{matrix}\right.\)\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
d) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{c}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Đẳng thức đầu tiên sai:
Ví dụ: \(a=1;b=2;c=3;d=6\) thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Nhưng \(\dfrac{a.d}{c.d}\ne\dfrac{a^2-b^2}{b^2-d^2}\)
Với đẳng thức thứ 2:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}+1=\dfrac{c}{d}+1=>\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}-1=\dfrac{c}{d}-1=>\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=>ad=cb=>ad+ac=cb+ac\)
\(=>a\left(c+d\right)=c\left(a+b\right)=>\dfrac{a}{c}=\dfrac{a+b}{c+d}=>\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
a: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
Cách 1:
Ta xét tích a(c-d) và c(a-b)
Ta có: a(c-d)=ac-ad (1)
c(a-b)=ac-bc(2)
Ta lại có \(\dfrac{a}{c}=\dfrac{c}{d}\)=>ad=bc (3)
Từ (1), (2), (3) ta có a(c-d)=c(a-d). Do đó \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cách 2:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k thì a=bk, c=dk.
Xét \(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
Xét \(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cách 3: Ta có
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Aps dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a-b}{c-d}\)
=>\(\dfrac{a}{c}=\dfrac{a-b}{c-d}=>\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
hay \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
C. \(\dfrac{b}{d}=\dfrac{c}{a}\)
Chúc bạn học tốt!!