Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh: \(\dfrac{3a^2+5ac}{3a^2-5ac}=\dfrac{3b^2+5bd}{3b^2-5bd}\)
Ai nhanh mình tick cho nhé
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Hãy chứng minh rằng :
\(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(\dfrac{a+2c}{b+2d}=\dfrac{a-2c}{b-2d}\)
\(\dfrac{a^2+2b^2}{c^2+2d^2}=\dfrac{a^2-2b^2}{c^2-2d^2}\)
\(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Cho \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{5a+3b}{5a-3b}\) = \(\dfrac{5c+3d}{5c+3d}\)
b, \(\dfrac{2a-b}{2a+b}\) = \(\dfrac{2c-d}{2c+d}\)
Cho:\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)(a,b,c,d > 0)tính giá trị của biểu thức P=\(\dfrac{2018a-2017b}{c+d}+\dfrac{2018b-2017}{a+d}+\dfrac{2018c-2017a}{b+c}+\dfrac{2018d-2017a}{b+c}\)
Cho dãy tỉ số bằng nhau:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
Tính : \(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
1. cho \(\dfrac{a}{b}\) = \(\dfrac{d}{c}\)
a) \(\dfrac{a+b}{b}\) = \(\dfrac{c+d}{d}\)
b) \(\dfrac{a}{a-b}\) = \(\dfrac{c}{c-d}\)
c) \(\dfrac{2a-5b}{2c-5d}\) = \(\dfrac{3a+4b}{3c+4d}\)
d) \(\dfrac{a^2-c^2}{b^2-d^2}\) = \(\dfrac{ac}{bd}\)
HELP ME !!! MK SẼ TICK CHO
Tìm a,b,c biết:
a. \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\); \(\dfrac{b}{5}\)=\(\dfrac{c}{4} \) và a-b+c=-49
b. 3a=2b; 5b=7c và 3a+5b-7c=60
c. \(\dfrac{a}{b}\)=\(\dfrac{8}{5}\), \(\dfrac{b}{3}=\dfrac{c}{5}\)và a+b+c=61
d.\(\dfrac{a}{3}=\dfrac{b}4, \dfrac{b}4=\dfrac{c}5\)và 2a-3b+c=6
e.\(\dfrac{a}2=\dfrac{b}3, \dfrac{b}4=\dfrac{c}5 và a^{2}-b^2=-16\)
các số a,b,c,d thỏa mãn điều kiện \(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}va\alpha+b+c+d\ne0\)
