Đặt  \(A=\frac{19^{19}-5}{19^{20}+4}\)

\(\Rightarrow19A=\frac{19^{20}-95}{19^{20}+4}=\frac{19^{20}+4-99}{19^{20}+4}=1-\frac{99}{19^{20}+4}\)

\(B=\frac{19^{20}-5}{19^{21}+4}\)

\(\Rightarrow19B=1-\frac{99}{19^{21}+4}\) ( chỗ này bn lm giống như mk ở trên nha! )

\(\Rightarrow\frac{99}{19^{20}+4}>\frac{99}{19^{21}+4}\)

\(\Rightarrow1-\frac{99}{19^{20}+4}< 1-\frac{99}{19^{21}+4}\)

\(\Rightarrow19A< 19B\)

=> A < B

Chọn đáp án:C

P=E=Z=19

N=A - Z= 39 - 19 = 20

=> Chọn C

C = D

olm duyệt đi

ca cach giai

mk tính đơn giản lắm chỉ cần tính hiệu của C va D thì ra hết..:))

Ta có: \(A=\frac {19^{20}+5}{19^{20}-8}=\frac {19^{20}-8+13}{19^{20}-8}=1+\frac {13}{19^{20}-8}\)

           \(B=\frac {19^{20}+6}{19^{20}-7}=\frac {19^{20}-7+13}{19^{20}-7}=1+\frac {13}{19^{20}-7}\)

Vì \(19^{20}-8<19^{20}-7\) nên \(\frac {13}{19^{20}-8}>\frac {13}{19^{20}-7}\)

\(\Rightarrow\)\(1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\) Hay \(A>B\) 

Vậy A>B

ta có A = \(\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\) 

và B = \(\frac{19^{20}+6}{19^{20}-7}=\frac{19^{20}-7+13}{19^{20}-7}=1+\frac{13}{19^{20}-7}\)

vì \(\frac{13}{19^{20}-8}>\frac{13}{19^{20}-7}\)\(\Rightarrow1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\)\(\Rightarrow A>B\)

Bài làm :

\(\left(19^{20}+19^{19}\right):19^{18}\)

\(=19^{20}:19^{18}+19^{19}:19^{18}\)

\(=19^2+19\)

\(=361+19\)

\(=380\)

Học tốt

\(\left(19^{20}+19^{19}\right)\div19^{18}\)

\(=19^2+19\)

\(=361+19\)

\(=380\)

\(\left(19^{21}+19^{22}+19^{23}\right):\left(19^{20}+19^{21}+19^{22}\right)\)

\(=19^{21}.\left(1+19+19^2\right):19^{20}:\left(1+19+19^2\right)=19\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}\left(19SH\right)\)

\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{20}>\frac{19}{20}\)

Vậy ................

Đặt \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\) ta có : 

\(A>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)

Do có \(20-2+1=19\) phân số \(\frac{1}{20}\) nên : 

\(A>19.\frac{1}{20}=\frac{19}{20}\)

Vậy \(A>\frac{19}{20}\)

Chúc bạn học tốt ~ 

\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)

\(=\dfrac{1}{20}\)