\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\)

\(=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\right)\)

Các phân số \(\dfrac{1}{5}\), \(\dfrac{1}{6}\), \(\dfrac{1}{7}\), ..., \(\dfrac{1}{19}\) đều lớn hơn \(\dfrac{1}{20}\), tất cả có 15 phân số nên:

\(B>\dfrac{1}{4}+\left(\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)

Vậy B > 1

e! Chung minh di tai sao lai lam the : phai co ly do chu( ko phai cu thich la ko lam ngay duoc dau

Đề có chuẩn ko vậy bn???

mk bít làm nhưng sai đề nè bạn

đề chuẩn ko đó

tóm lại là bạn gõ sai đề rùi

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

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

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

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)

\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)

B= 1/4+(1/5+1/6+...+1/9)+(1/10+1/11+...+1/19)
Vì 1/5+1/6+...+1/9 > 1/9+1/9+...+1/9 nên 1/5+1/6+...+1/9 > 5/9 >1/2
Vì 1/10+1/11+...+1/19 > 1/19+1/19+...+1/19 nên 1/10+1/11+...+1/19 > 10/19 >1/2
Suy ra: B > 1/4+1/2+1/2 > 1

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...............

\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)

=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)

B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))

Vậy.....

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)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B< 1-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

mink nhanh nhất đó bạn,

ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1\times2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\times3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\times4}\)

. . . . . . .

\(\dfrac{1}{8^2}< \dfrac{1}{7\times8}\)

_________________________________

\(\Rightarrow\)\(B< \)\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{7}-\dfrac{1}{8}\)

\(\Rightarrow B< 1-\dfrac{1}{8}\)

\(\Rightarrow B< 1\)

\(\Rightarrowđpcm\)