Bài 8: Tính chất của dãy tỉ số bằng nhau
Các câu hỏi tương tự
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 \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
18 tháng 10 2021 lúc 13:25
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 \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{c}{d}\). chứng mk: \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
12 tháng 11 2021 lúc 21:16
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
1. Cho dfrac{a}{b} dfrac{c}{d} c/m
a) (2a+3c) . (2b-3d) (2a- 3c) . (2b+3d)
b) dfrac{left(a^2+cright)^2}{left(b+dright)^2} dfrac{left(a-cright)^2}{left(b-dright)^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 ~ !!!
Đọc tiếp
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 >~< !!!
cho\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\).CTR\(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}\)=\(\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}\)
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
15 tháng 2 2018 lúc 9:07
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) CMR :\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(cho\)cho:\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}.\) cmr: \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)