Ta có \(\frac{c}{d}>\frac{a}{b}<=>cb>ad \)
<=>bc+cd>ad+cd
<=>c(b+d)>d(a+c)
<=>\(\frac{c}{d}>\frac{a+c}{b+d} \)
cmtt =>\(\frac{a+c}{b+d}>\frac{a}{b} \)
cho tỉ lệ thức\(\dfrac{a}{b}=\dfrac{c}{d}\)
(a,b,c,d khác 0)
chứng tỏ rằng
bài 1 \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
bài 2 \(\dfrac{2a+c}{3a-c}=\dfrac{2b+d}{3b-d}\)
bài 3\(\dfrac{5a-2c}{3a-4c}=\dfrac{5b-2d}{3b-4d}\)
nhanh nha gấp lắm ạ
Cho ad = bc và a, b, c, d khác 0. Chứng tỏ rằng:
a) \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) b) \(\dfrac{a+c}{b+d}\) = \(\dfrac{a}{b}\) c) \(\dfrac{a}{c}\) = \(\dfrac{b}{d}\) d) \(\dfrac{a+b}{b}\) = \(\dfrac{c+d}{d}\) e) \(\dfrac{2a+b}{2c+d}\) = \(\dfrac{a}{c}\)
Chứng minh rằng :
Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\) (với a,b,c,d \(\ne\) 0)
cho a+b+c+d khác 0 vàti\(\dfrac{b+c+d-a}{a}=\dfrac{c+d+a-b}{b}=\dfrac{d+a+b-c}{c}=\dfrac{a+b+c-d}{d}P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{c}{d}\right)\left(1+\dfrac{a}{d}\right)\)tính P
giúp mk với ạ , xin cảm ơn
Cho \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}\)
CMR \(\dfrac{a^3+c^3+m^3}{b^3-d^3-n^3}\) = \(\left(\dfrac{a+c-m}{b+d-m}\right)^3\)
mọi người ơi giup mik với ai làm đc mik tick cho
Cho\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) với b+c+d khác 0.
Chứng minh:\(\dfrac{a^3+b^3+c^3}{b^3+c^3-d^3}=\left(\dfrac{a+d-c}{b+c-d}\right)^3\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{c}{d}\). chứng mk: \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
1. Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
2. Cho \(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}\). Chứng minh rằng \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Chứng minh rằng :
Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\ne1\) thì \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) với \(a,b,c,d\ne0\)
