Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b.\left(3k+1\right)}=\dfrac{k}{3k+1}\)(1)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d.\left(3k+1\right)}=\dfrac{k}{3k+1}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\) (đpcm)
Chúc bạn học tốt!!!
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}.\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\left(1\right)\)
Thay (1) vào đề bài:
\(VT=\dfrac{a}{3a+b}=\dfrac{ck}{3ck+dk}=\dfrac{ck}{k\left(3c+d\right)}=\dfrac{c}{3c+d}\)
\(VP=\dfrac{c}{3c+d}=VT\)
\(\Leftrightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\rightarrowĐPCM.\)
\(\text{Đặt}\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left[{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\text{Ta có:}\)
\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\) \(\left(1\right)\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\) \(\left(2\right)\)
\(Từ\) \(\left(1\right)\) \(và\) \(\left(2\right)\) \(suy\) \(ra:\)
\(\dfrac{a}{3a+b}=\dfrac{k}{3k+1}=\dfrac{c}{3c+d}\)
\(Vậy\) \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\) \(\left(ĐPCM\right)\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)⇒\(\dfrac{a}{c}=\dfrac{b}{d}\)
⇒\(\dfrac{a}{c}=\dfrac{3a}{3c}=\dfrac{b}{d}=\dfrac{3a+b}{3c+d}\)⇒\(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
Vậy nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì\(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
