=\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=\(\dfrac{d}{5}\)Và a +b +c + d =-42

\(\dfrac{a}{2}\) = \(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=\(\dfrac{d}{5}\) =\(\dfrac{a+b+c+d}{2+3+4+5}\)= \(\dfrac{-42}{14}\)= 3

Vậy A=6,B=9,C=12,D=15

\(\Rightarrow\)\(\dfrac{a}{2}\) =\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=\(\dfrac{d}{5}\) Và a +b +c + d =-42

\(\dfrac{a}{2}\) = \(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)= \(\dfrac{d}{5}\)=\(\dfrac{a+b+c+d}{2+3+4+5}\)=\(\dfrac{-42}{14}\)=-3

\(\dfrac{ab}{ae}\)+\(\dfrac{ac}{af}\)=3

Ta có 3a + b =105 + 2c

\(\Rightarrow\) 3a + b - 2c + 4d =105

\(\Rightarrow\)\(\dfrac{a}{2}\)= \(\dfrac{b}{3}\)= \(\dfrac{c}{4}\) = \(\dfrac{d}{5}\)

\(\Rightarrow\)\(\dfrac{3a}{6}\)= \(\dfrac{b}{3}\)=\(\dfrac{2c}{8}\) =\(\dfrac{4d}{20}\)= \(\dfrac{3a+b-2c+4d}{6+3-8+20}\)=\(\dfrac{105}{21}\)=\(5\)

\(\Rightarrow\)a= 5x6:3=10

\(\Rightarrow\)b=5x3=15

\(\Rightarrow\)c=5x8:2=20

\(\Rightarrow\)d=5x20:4=25

Ta có \(2^{90}\) =(\(2^5\))\(^{18}\)= \(32^{18}\)

\(5^{36}\) =(\(5^2\))\(^{18}\) =\(25^{18}\)

Vì 32 > 25 nên \(32^{18}\)>\(25^{18}\)\(\Rightarrow\)\(2^{90}\)>\(5^{36}\)

Vậy \(2^{90}\)>\(5^{36}\)

Ta có \(2^{90}\) = ( \(2^{5_{ }^{ }}\) )