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:\(\dfrac{a+\overline{bc}}{\overline{abc}}=\dfrac{b+\overline{ca}}{\overline{bca}}=\dfrac{c+\overline{ab}}{\overline{cab}}\)
CMR:\(\overline{\dfrac{bc}{a}=\dfrac{\overline{ca}}{b}=\dfrac{\overline{ab}}{c}}\)
Cho 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 \(\dfrac{\overline{abc}}{a+\overline{bc}}=\dfrac{\overline{bca}}{b+\overline{ca}}\)
CM: \(\dfrac{a}{\overline{bc}}=\dfrac{b}{\overline{ca}}\)
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
Bài 1: Tìm 2 số dương biết rằng tổng, hiệu, tích của chúng tỉ lệ với \(\dfrac{1}{30},\dfrac{1}{120},\dfrac{1}{16}\).
Bài 2: Tìm k, biết:
\(k=\dfrac{\overline{ab}}{\overline{abc}}=\dfrac{\overline{bc}}{\overline{bca}}=\dfrac{\overline{ca}}{\overline{cab}}\)
Cho:\(\dfrac{\overline{ab}}{b}=\dfrac{\overline{bc}}{c}=\dfrac{\overline{ca}}{a}\)
CMR(\(\overline{abc}\))123=111123\(\cdot a^{40}\cdot b^{41}\cdot c^{42}\)
Cho \(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}\). Chứng minh rằng \(\dfrac{\overline{abbb}}{\overline{bbbc}}=\dfrac{a}{c}\)
