Ta có :
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
Vì \(1+\dfrac{2}{20^{10}-1}< 1+\dfrac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
~ Chúc bn học tốt~
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\) (1)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\) (2)
vì \(20^{10}-1>20^{10}-3\)
nên \(\dfrac{2}{20^{10}-1}< \dfrac{2}{20^{10}-3}\) (3)
từ (1), (2) và (3) suy ra A<B
vì B>1
Suy ra B>\(\dfrac{20^{10}-1+2}{20^{10}-3+2}=\dfrac{20^{10}+1}{20^{10}-1}=A\)
Suy ra B>A
Ta co : A = 20^10+1/20^10-1= 20^10+2-1/20^10-1=20^10-1/20^10-1+2/20^10-1=1+2/20^10-1
B = 20^-1/20^10-3= 20^10-3+2/20^10-3= 20^10-3/20^10-3+2/20^10-3= 1+2/20^10-3
Vi 20^10-1>20^10-3(ma mau cang be thi phan so cang lon, mau cang lon thi phan so cang be)nén+2/20^10-1<1+2/20^10-3 hay A<B
Vay A<B
