a² + b² + c² + d² – ab – ac – ad = 0 (1)
 Nhân hai vế của với 4 rồi đưa về dạng : a² + (a – 2b)² + (a – 2c)² +(a – 2d)² = 0 (2). Do đó ta có :
a = a – 2b = a – 2c = a – 2d = 0 . Suy ra : a = b = c = d = 0.

a+b+c+d=0

Vì a/b < c/d (Với a,b,c,d thuộc N*)

=> ad<bc

=>  2018ad < 2018bc

=> 2018ad + cd < 2018bc +cd

=> (2018a + c).d < (2018b+d).c

=> 2018a +c / 2018b + d < c/d

lho1 quá ,bỏ qua nhé bạn!

\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)

\(\Rightarrow2019ad< 2019bc\)

\(\Rightarrow2019ad+cd< 2019bc+cd\)

\(\Rightarrow d\left(2019a+c\right)< c\left(2019b+d\right)\)

\(\Rightarrow\frac{2019a+c}{2019b+d}< \frac{c}{d}\left(đpcm\right)\)

Ta có \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)

\(\Leftrightarrow2018ad< 2018bc\)

\(\Leftrightarrow2018ad+cd< 2018bc+cd\)

\(\Leftrightarrow d\left(2018a+c\right)< c\left(2018b+d\right)\)

\(\Rightarrow\frac{2018a+c}{2018b+d}< \frac{c}{d}\left(đpcm\right)\)

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

\(\Leftrightarrow\left(a+b+c\right)\left(a-b-c\right)=\left(a-b+c\right)\left(a+b-c\right)\)\(\Leftrightarrow a^2-ab-ac+ab-b^2-bc+ac-bc-c^2=a^2-ab+ac+ab-b^2+bc-ac+bc-c^2\)

\(\Leftrightarrow4bc=0\) \(\Leftrightarrow bc=0\)

\(\Rightarrow D=0\)

\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3a+3b+3c+3d}=\frac{1}{3}.\) (T/c dãy tỷ số bằng nhau)

=> \(\frac{a}{3b}=\frac{1}{3}\Rightarrow\frac{a}{b}=1\Rightarrow a=b\)

Làm tương tự sẽ rút ra a=b=c=d