Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3b+3c+3d+3x}=\dfrac{a+b+c+d}{3.\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow a=\dfrac{1}{3}.3b=b\\ \Rightarrow b=\dfrac{1}{3}.3c=c\\ \Rightarrow c=\dfrac{1}{3}.3d=d\\ \Rightarrow d=\dfrac{1}{3}.3a=a\)
➩\(\text{a=b=c=d}\)
Tick cho mình nhé
Cho các số a, b, c, d thõa mản điều kiện:
\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}\) và \(a+b+c+d\ne0\)
CMR: a = b = c = d
Cho a,b,c là 3 số thực khác 0, thỏa mãn điều kiện:
\(\dfrac{a+b-c}{3a}=\dfrac{b+c-a}{3b}=\dfrac{c+a-b}{3c}\)
Tìm a,b,c
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 \(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{1}{a+b+c}\left(a,b,c\ne0,a+b+c\ne0\right)\)
Tính \(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)\)
Ai giúp mik đi, mik cho 5 coin
cho \(\dfrac{a}{b}=\dfrac{c}{d}\) khác \(\pm\)1 và c khác 0 , CMR
a) \(\left(\dfrac{a-b}{c-d}\right)^2=\dfrac{ab}{cd}\)
b)\(\dfrac{5a+3b}{5c+3d}\) =\(\dfrac{5a+3b}{5c+3d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.\)Chứng minh \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
Cho \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) . Chứng minh :
a, \(\dfrac{5a+3b}{5a-3b}\) = \(\dfrac{5c+3d}{5c+3d}\)
b, \(\dfrac{2a-b}{2a+b}\) = \(\dfrac{2c-d}{2c+d}\)
1. Cho \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) c/m
a) (2a+3c) . (2b-3d) = (2a- 3c) . (2b+3d)
b) \(\dfrac{\left(a^2+c\right)^2}{\left(b+d\right)^2}\) = \(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)
c)\(\dfrac{a^3+b^3}{c^3+d^3}\) = \(\dfrac{a^3-b^3}{c^3-d^3}\)
d) \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}\) = \(\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}\)
HELP ME >~< !!!
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
Chứng minh: \(\dfrac{a}{b}=\dfrac{c}{d}\)
