Thay a + c = 2b vào 2bd = c(b+d), ta có :
\(\left(a+c\right)d=c\left(b+d\right)\\ \Leftrightarrow\dfrac{a+c}{c}=\dfrac{b+d}{d}\\ \Leftrightarrow\dfrac{a}{c}+1=\dfrac{b}{d}+d\\ \Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\\ \dfrac{a}{b}=\dfrac{c}{d}\\ \RightarrowĐPCM\)
Chứng minh rằng nếu a + c = 2b và 2bd = c.(b + d) với b, d khác 0 thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Chứng minh rằng nếu a+c=2b và 2bd = c (b+d) thì \(\dfrac{a}{b}=\dfrac{c}{d}\) với b , d khác 0
Nếu a+c=2b và 2bd=c(b+d) thì \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\left(b,d\ne0\right)\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
b, \(\dfrac{c}{a+c}=\dfrac{b}{b+d}\)
c, \(\dfrac{a+b}{a}=\dfrac{d}{c+d}\)
d, \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
e, \(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
f, \(\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{c^2+d^2}{c^2-d^2}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\).CMR:
a) \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
b) \(\dfrac{a-b}{c-d}=\dfrac{a^2-b^2}{c^2-d^2}\)
c) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
d) \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{2b^2+5bd}{7b^2-5bd}\)
Cho
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
Tính:
\(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+4c}{b+4d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a+2c}{3b+2d}\)
c) \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-2b}{c-2d}\)
d) \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a-2b}{5c-2d}\)
Cho dãy tỉ 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 \(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
Bài 1
Cho \(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
Tính M = \(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
