Toán Cho $\frac{a}{b}$ = $\frac{c}{d}$ . CMR $\frac{a}{3a + b}$ = $\frac{c}{3c + d}$ 13/11/2021 By Madeline Cho $\frac{a}{b}$ = $\frac{c}{d}$ . CMR $\frac{a}{3a + b}$ = $\frac{c}{3c + d}$
Đặt $\dfrac ab =\dfrac cd = k\quad (k\ne 0)$ $\to \begin{cases}a = kb\\c = kd\end{cases}$ Ta được: $+)\dfrac{a}{3a +b}=\dfrac{kb}{3kb + b}=\dfrac{kb}{b(3k + 1)}=\dfrac{k}{3k+1}$ $+)\dfrac{c}{3c+d}=\dfrac{kd}{3kd+d}=\dfrac{kd}{d(3k+1)}=\dfrac{k}{3k+1}$ Do đó: $\dfrac{a}{3a +b}=\dfrac{c}{3c+d}$ Trả lời
Đặt `a/b = c/d = k => a=bk ; c=dk` `=> a/(3a+b) = (bk)/(3bk+b) = (bk)/(b.(3k+1)) = k/(3k+1)` `=> c/(3c+d) = (dk)/(3dk+d) = (dk)/(d.(3k+1)) = k/(3k+1)` `=> a/(3a+b) = c/(3c+d) (đpcm)` XIN HAY NHẤT Ạ Trả lời
Đặt $\dfrac ab =\dfrac cd = k\quad (k\ne 0)$
$\to \begin{cases}a = kb\\c = kd\end{cases}$
Ta được:
$+)\dfrac{a}{3a +b}=\dfrac{kb}{3kb + b}=\dfrac{kb}{b(3k + 1)}=\dfrac{k}{3k+1}$
$+)\dfrac{c}{3c+d}=\dfrac{kd}{3kd+d}=\dfrac{kd}{d(3k+1)}=\dfrac{k}{3k+1}$
Do đó:
$\dfrac{a}{3a +b}=\dfrac{c}{3c+d}$
Đặt `a/b = c/d = k => a=bk ; c=dk`
`=> a/(3a+b) = (bk)/(3bk+b) = (bk)/(b.(3k+1)) = k/(3k+1)`
`=> c/(3c+d) = (dk)/(3dk+d) = (dk)/(d.(3k+1)) = k/(3k+1)`
`=> a/(3a+b) = c/(3c+d) (đpcm)`
XIN HAY NHẤT Ạ