Question

[You Are Mine]

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 >~< !!!

Answer 1

[Nhan Mạc Oa]

a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)

VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)

Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)

Answer 2

[You Are Mine]

thanks bn nhìu nha