Ta có:
\(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}=\dfrac{9a+b}{100b}=\dfrac{999a+111b}{1110b}=\dfrac{999a+a+111b}{1110b}=\dfrac{1000a+111b}{1110b+c}=\dfrac{\overline{abbb}}{\overline{bbbc}}\)
\(\Rightarrowđpcm\)
Chúc bạn học tốt!
Ta có:
\(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}=\dfrac{9a+b}{100b}=\dfrac{999a+111b}{1110b}=\dfrac{999a+a+111b}{1110b}=\dfrac{1000a+111b}{1110b+c}=\dfrac{\overline{abbb}}{\overline{bbbc}}\)
\(\Rightarrowđpcm\)
Chúc bạn học tốt!
Cho \(\dfrac{a+\overline{bc}}{\overline{abc}}=\dfrac{b+\overline{ca}}{\overline{bca}}=\dfrac{c+\overline{ab}}{\overline{cab}}\). Chứng minh rằng \(\dfrac{\overline{ab}}{c}=\dfrac{\overline{ca}}{b}=\dfrac{\overline{bc}}{a}\)
Cho \(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}\). Cmr : \(\dfrac{\overline{abbb...b}}{\overline{bbb...bc}}=\dfrac{a}{c}\)
( n-1 chữ số b ở cả tử và mẫu )
Cho tỉ lệ thức \(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}.\)CMR \(\dfrac{\overline{abbbb...bbb}}{\overline{bbbb...bbbc}}=\dfrac{a}{c}\)( có n chữ số b; n là số tự nhiên)
Cho:\(\dfrac{a+\overline{bc}}{\overline{abc}}=\dfrac{b+\overline{ca}}{\overline{bca}}=\dfrac{c+\overline{ab}}{\overline{cab}}\)
CMR:\(\overline{\dfrac{bc}{a}=\dfrac{\overline{ca}}{b}=\dfrac{\overline{ab}}{c}}\)
cho \(\dfrac{\overline{abc}}{\overline{bc}}=\dfrac{\overline{bca}}{\overline{ca}}=\dfrac{\overline{cab}}{\overline{ab}}\). Tính \(\dfrac{a}{\overline{bc}}+\dfrac{b}{\overline{ca}}+\dfrac{c}{\overline{ab}}\)
Cho \(\dfrac{\overline{ab}+\overline{bc}}{a+c}=\dfrac{\overline{bc}+\overline{ca}}{b+c}=\dfrac{\overline{ca}+\overline{ab}}{c+a}\)
CMR : a = b = c
Cho biết \(\dfrac{\overline{abc}}{\overline{bc}}=\dfrac{\overline{bca}}{\overline{ca}}=\dfrac{\overline{cab}}{\overline{ab}}\)
Tính tổng\(\dfrac{a}{\overline{bc}}+\dfrac{b}{\overline{ca}}+\dfrac{c}{\overline{ab}}\)
Cho \(\dfrac{\overline{abc}}{a+\overline{bc}}=\dfrac{\overline{bca}}{b+\overline{ca}}\). Chứng minh: \(\dfrac{a}{\overline{bc}}=\dfrac{b}{\overline{ca}}\)
Cho tỉ lệ thức \(\dfrac{\overline{ab}}{\overline{bc}}\) = \(\dfrac{b}{c}\) (c\(\ne\) 0). Chứng minh rằng \(\dfrac{a^2+b^2}{b^2+c^2}\) = \(\dfrac{a}{c}\)