c: \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
mà \(-\frac{9}{10^{12}-1}<0<\frac{9}{10^{11}+1}\)
nên 10A<10B
=>A<B
d: \(A=\frac{5\cdot\left(11\cdot13-22\cdot26\right)}{22\cdot26-44\cdot52}\)
\(=\frac{5\cdot\left(11\cdot13-22\cdot26\right)}{4\left(11\cdot13-22\cdot26\right)}=\frac54=1+\frac14\)
\(B=\frac{138^2-690}{137^2-548}=\frac{138\cdot\left(138-5\right)}{137\cdot\left(137-4\right)}=\frac{138}{137}=1+\frac{1}{137}\)
4<137
=>\(\frac14>\frac{1}{137}\)
=>\(\frac14+1>\frac{1}{137}+1\)
=>A>B
