Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/b=c/d
=> ad=bc
=>ac-ad=ac-bc
=>a(c-d)=c(a-b)
=> a/(a-b)=c/(c-d)
Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}\)
\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1\)
\(\Rightarrow\frac{b}{a}-\frac{a}{a}=\frac{d}{c}-\frac{c}{c}.\)
\(\Rightarrow\frac{b-a}{a}=\frac{d-c}{c}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\left(đpcm\right).\)
Chúc bạn học tốt!
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\left(1\right)\\ \Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\frac{a-b}{c-d}=\frac{ck-dk}{c-d}=\frac{k\left(c-d\right)}{c-d}=k\left(2\right)\)
(1)(2) \(\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\)
Đặt\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=k\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=k\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=k^{2013}\)(1)
Mặt khác:\(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=k^{2013}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}=k^{2013}\)(2)
Từ (1);(2) ta có: \(\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(=k^{2013}\right)\)
có \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)=>\(\frac{a^{2013}}{c^{2013}}=\frac{\left(a-b\right)^{2013}}{\left(c-d\right)^{2013}}\)
ngược lại cũng có \(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)
=> đpcm :V
ta có:\(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{b}{a}=\frac{c}{d}\)
\(\Rightarrow1-\frac{b}{a}=1-\frac{c}{d}\)
\(\Rightarrow\frac{a}{a}-\frac{b}{a}=\frac{c}{c}-\frac{d}{c}\)
\(\Rightarrow\frac{a-b}{a}=\frac{c-d}{c}\)
hay: \(\frac{a}{a-b}=\frac{c}{c-d}\)(đpcm)
Cách 1 : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
Cách 2 : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ac-ad=ac-bc\)
\(\Rightarrow a(c-d)=c(a-b)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
Cách 3 : Đặt \(\frac{a}{b}=\frac{c}{d}=m\Rightarrow a=mb,c=md\)
Ta có : \(\frac{a}{a-b}=\frac{mb}{mb-b}=\frac{mb}{b(m-1)}=\frac{m}{m-1}\)
\(\frac{c}{c-d}=\frac{md}{md-d}=\frac{md}{d(m-1)}=\frac{m}{m-1}\)
Do đó : \(\frac{a}{a-b}=\frac{c}{c-d}\)
Cách 4 : \(\frac{a}{a-b}=\frac{c}{c-d}\Rightarrow a(c-d)=c(a-b)\)
\(\Rightarrow ac-ad=ac-bc\Rightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\) đẳng thức đúng
Do đó , ta có : \(\frac{a}{a-b}=\frac{c}{c-d}\)là đẳng thức đúng.
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có
\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)
\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)
\(\Rightarrow VT=VP\)
Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)
a) \(\frac{a}{a+b}=\frac{c}{c+d}\)=> a . ( c + d ) = c . ( a + b )
=> ac + ad = ac + cb
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\)
Ta có: a/b=c/d =>a.d=b.c
a/a-b=a.d/d.(a-b)=b.c/a.d-b.d=b.c/b.c-b.d=b.c/b.(c-d)=c/c-d
<=>a/a-b=c/c-d(ĐPCM)
Gọi \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=kb;c=kd\)
Thay vào ta có :
\(\frac{a}{a-b}=\frac{kb}{kb-b}=\frac{kb}{\left(k-1\right)b}=\frac{k}{k-1}\)
\(\frac{c}{c-d}=\frac{kd}{kd-d}=\frac{kd}{\left(k-1\right)d}=\frac{k}{k-1}\)
Mà \(\frac{k}{k-1}=\frac{k}{k-1}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
\(\RightarrowĐPCM\)