Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\dfrac{b^2k^{2022}+b^{2022}}{d^{2022}k^{2022}+d^{2022}}=\left(\dfrac{b}{d}\right)^{2022}\)

\(\dfrac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}=\dfrac{\left(bk+b\right)^{2022}}{\left(dk+d\right)^{2022}}=\left(\dfrac{b}{d}\right)^{2022}\)

=>\(\dfrac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\dfrac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\frac{\left(bk\right)^{2022}+b^{2022}}{\left(dk\right)^{2022}+d^{2022}}=\frac{b^{2022}\left(k^{2022}+1\right)}{d^{2022}\left(k^{2022}+1\right)}=\frac{b^{2022}}{d^{2022}}\)

\(\frac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}=\frac{\left(bk+b\right)^{2022}}{\left(dk+d\right)^{2022}}=\frac{\left\lbrack b\left(k+1\right)\right\rbrack^{2022}}{\left\lbrack d\left(k+1\right)\right\rbrack^{2022}}=\frac{b^{2022}}{d^{2022}}\)

Do đó: \(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\frac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}\)

a^2+b^2+c^2=ab+bc+ac

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0

=>(a-b)^2+(b-c)^2+(a-c)^2=0

=>a=b=c

\(T=\dfrac{a^{2022}+a^{2022}+a^{2022}}{\left(3a\right)^{2022}}=\dfrac{3}{3^{2022}}=\dfrac{1}{3^{2021}}\)

\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)

\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)

\(\Rightarrow A< B\)

TA có: \(3a-b=\frac12\left(a+b\right)\)

=>2(3a-b)=a+b

=>6a-2b=a+b

=>5a=3b

=>a=0,6b

\(C=\frac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)

\(=\frac{\left(0,6b\right)^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)

\(=\frac{\left(0,6\right)^{2022}\left(b^{2022}+5^{2022}\right)}{b^{2022}+5^{2022}}=\left(0,6\right)^{2022}=\left(\frac35\right)^{2022}\)

3a-b=1/2(a+b)

=>6a-2b=a+b

=>5a=3b

=>a/3=b/5=k

=>a=3k; b=5k

\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)

\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)