Bài 18:
$\dfrac{b+c+d}{a} = \dfrac{c+d+a}{b} = \dfrac{d+a+b}{c} = \dfrac{a+b+c}{d} \\
= \dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d} \\
= \dfrac{3a+3b+3c+3d}{a+b+c+d} \\
= \dfrac{3(a+b+c+d)}{a+b+c+d} = 3$
Vậy $k=3$
Bài 19:
Đặt $\dfrac{a}{2}=\dfrac{b}{5} = \dfrac{c}{7}=k \ (k \in \mathbb{R}, \ k \neq 0) \\
\Rightarrow a=2k,\ b=5k, \ c=7k$
$A=\dfrac{a-b+c}{a+2b-c} = \dfrac{2k-5k+7k}{2k+2.5k-7k}= \dfrac{k(2-5+7)}{k(2+10-7)} = \dfrac{4}{5}$
Bài 21: + $\dfrac{a}{2b} = \dfrac{b}{2c} = \dfrac{c}{2d} = \dfrac{d}{2a} = \dfrac{a+b+c+d}{2(a+b+c+d)} = \dfrac{1}{2}$
$\Rightarrow a = \dfrac{2b}{2}=b, \ b = \dfrac{2c}{2}=c, \ c = \dfrac{2d}{2}=d, \ d = \dfrac{2a}{2}=a$
$\Rightarrow a=b=c=d$
+ Ta có: $A= \dfrac{2021a-2022b}{c+d} + \dfrac{2021b-2022c}{a+d} + \dfrac{2021c-2022d}{a+b} +\dfrac{2021d-2022a}{b+c} \\
= \dfrac{2021a-2022a}{2a} + \dfrac{2021a-2022a}{2a} + \dfrac{2021a-2022a}{2a} + \dfrac{2021a-2022a}{2a} \\
= \dfrac{-4a}{2a}=-2$