a) \(\dfrac{a}{b}=\dfrac{c}{d}\)suy ra\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
ta có \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
nên \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b)đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) suy ra a=bk;c=dk
ta có \(\dfrac{a}{b+a}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)(1)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)(2)
Từ (1);(2) suy ra \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c)ta có \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
suy ra \(\dfrac{2a}{2c}=\dfrac{5b}{5d};\dfrac{3a}{3c}=\dfrac{4b}{4d}\)
suy ra \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3a-4d}\)
nên \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+3d}{3c-4d}\)
d)\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2\left(1\right)\)\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\left(2\right)\)
từ (1);(2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
