Lời giải:

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

A.2=4/1.5+6/5.11+...+12/29.41

A.2=1-1/5+1/5-1/11+...+1/29-1/41

A.2=1-1/41

A.2=40/41

A=20/41

B.3=3/1.4+6/4.10+...+12/29.31

B.3=1-1/4+1/4-1/10+...+1/29-1/31

B.3=1-1/31

B.3=30/31

B=10/31

Vì 20/41.10/31 nên A>B

\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(\Rightarrow2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)

\(\Rightarrow2A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)

\(\Rightarrow2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(\Rightarrow A=\dfrac{40}{41}:2=\dfrac{20}{41}\)(1)

\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(\Rightarrow3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\)

\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{31}=\dfrac{30}{31}\)

\(\Rightarrow B=\dfrac{30}{31}:3=\dfrac{10}{31}\)

\(\Rightarrow B=\dfrac{2}{2}.\dfrac{10}{31}=\dfrac{20}{62}\)

+)Ta có:\(\dfrac{20}{62}< \dfrac{20}{41}\Rightarrow B< A\)

Hay A>B(ĐPCM)

Chúc bn học tốt

Giải:

\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\) 

\(2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\) 

\(2A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\) 

\(2A=\dfrac{1}{1}-\dfrac{1}{41}\) 

\(2A=\dfrac{40}{41}\) 

\(A=\dfrac{40}{41}:2\) 

\(A=\dfrac{20}{41}\) 

\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\) 

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\) 

\(3B=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\) 

\(3B=\dfrac{1}{10}-\dfrac{1}{31}\) 

\(3B=\dfrac{21}{310}\) 

\(B=\dfrac{21}{310}:3\) 

\(B=\dfrac{7}{310}\) 

Vì \(\dfrac{20}{41}>\dfrac{7}{310}\) nên A>B

a) \(\dfrac{23}{24}< 1\)

\(\dfrac{24}{23}>1\)

\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)

b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)

c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)

d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)

Lời giải:

a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)

Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$

Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$

Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$

b) Rõ ràng $10^{11}-1< 10^{12}-1$. 

Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$

Áp dụng kết quả phần a:

$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$

a)\(\dfrac{-36}{63};\dfrac{56}{63};\dfrac{-30}{63}\)

b)\(\dfrac{110}{264};\dfrac{21}{264}\)

a) Mẫu số chung là BCNN(7, 9, 21) = 3².7 = 63

Thừa số phụ của 7 là 9, của 9 la 7, của 21 là 3. Do đó:

b) Mẫu số chung: 2³.3.11 = 264. Do đó:

\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)

\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)

\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)

\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)

\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)

mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)

\(\Rightarrow A< B\)

       A =      \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)

     3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)

3A - A =  1 - \(\dfrac{1}{3^{2023}}\)

   2A   = 1 - \(\dfrac{1}{3^{2023}}\) < 1

      B =  \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)

      B  = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)

     B   = \(\dfrac{8}{12}\)

     B   = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1 

2A < 2B ⇒ A < B 

a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)

\(\dfrac{-5}{6}>\dfrac{-91}{104}\)

Sao bn giống BT mình thế ?:)

a) \(< \)

b) \(>\)

c) \(< \)

d) \(>\)

e) \(< \)

g) \(>\)

h) \(>\)

k) \(>\)