CMR: Nếu $\frac{a+b}{b+c}=\frac{c+d}{d+a} (c+d\neq0) thì $a=c$ hoặc $a=b+c+d=0$ 08/07/2021 Bởi Caroline CMR: Nếu $\frac{a+b}{b+c}=\frac{c+d}{d+a} (c+d\neq0) thì $a=c$ hoặc $a=b+c+d=0$
Lời giải: Theo bài ra ta có: $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$ $=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}$ $=\dfrac{a+b}{c+d}+1=\dfrac{b+c}{d+a}+1$ $=\dfrac{a+b+c+d}{c+d}=\dfrac{a+b+c+d}{d+a}$ $(a+b+c+d)(\dfrac{1}{c+d} – \dfrac{1}{d+a})=0$ `=>` \(\left[ \begin{array}{l}a+b+c+d=0\\\dfrac{1}{c+d} – \dfrac{1}{d+a}=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}a+b+c+d=0\\\ a=c\end{array} \right.\) Vậy ta có điều phải chứng minh. Bình luận
Lời giải:
Theo bài ra ta có:
$\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$
$=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}$
$=\dfrac{a+b}{c+d}+1=\dfrac{b+c}{d+a}+1$
$=\dfrac{a+b+c+d}{c+d}=\dfrac{a+b+c+d}{d+a}$
$(a+b+c+d)(\dfrac{1}{c+d} – \dfrac{1}{d+a})=0$
`=>` \(\left[ \begin{array}{l}a+b+c+d=0\\\dfrac{1}{c+d} – \dfrac{1}{d+a}=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}a+b+c+d=0\\\ a=c\end{array} \right.\)
Vậy ta có điều phải chứng minh.