Question

1. Cho tỉ lệ thức ��ba​ = ��dc​. CMR:

a) 3�+5�3�+5�3b+5d3a+5c​ = �−2��−2�b−2da−2c​.

b) �2−�2��aba2−b2​ = �2−�2��cdc2−d2​.

c) (�+�)2�2+�2a2+b2(a+b)2​ = (�+�)2�2+�2c2+d2(c+d)2​.

d) (�+��+�)3(c+da+b​)3 =  �3+�3�3+�3c3+d3a3+b3​.

Gíup mình với cảm ơn các bạn rất nhiều!!!!!!!!!

Answer 1

a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{3b+5d}{3a+5c}=\dfrac{3b+5d}{3bk+3dk}=\dfrac{1}{k}\)

\(\dfrac{b-2d}{a-2c}=\dfrac{b-2d}{bk-2dk}=\dfrac{1}{k}\)

=>\(\dfrac{3b+5d}{3a+5c}=\dfrac{b-2d}{a-2c}\)

b: \(\dfrac{ab}{a^2-b^2}=\dfrac{bk\cdot b}{b^2k^2-b^2}=\dfrac{k}{k^2-1}\)

\(\dfrac{cd}{c^2-d^2}=\dfrac{dk\cdot d}{d^2k^2-d^2}=\dfrac{k}{k^2-1}\)

=>ab/a^2-b^2=cd/c^2-d^2

c: \(\dfrac{a^2+b^2}{\left(a+b\right)^2}=\dfrac{b^2k^2+b^2}{\left(bk+b\right)^2}=\dfrac{k^2+1}{\left(k+1\right)^2}\)

\(\dfrac{c^2+d^2}{\left(c+d\right)^2}=\dfrac{d^2k^2+d^2}{\left(dk+d\right)^2}=\dfrac{k^2+1}{\left(k+1\right)^2}\)

=>\(\dfrac{a^2+b^2}{\left(a+b\right)^2}=\dfrac{c^2+d^2}{\left(c+d\right)^2}\)