Vì \(\frac{1}{201}>\frac{1}{400}\)

\(\frac{1}{202}>\frac{1}{400}\)

\(\frac{1}{203}>\frac{1}{400}\)

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

\(\frac{1}{399}>\frac{1}{400}\)

⇒ \(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{399}>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)(199 số hạng \(\frac{1}{400}\))

⇒ \(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{399}+\frac{1}{400}>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)(200 số hạng \(\frac{1}{400}\)) = 200.\(\frac{1}{400}\)=\(\frac{1}{2}\)

⇒ A > \(\frac{1}{2}\)

Vậy A > \(\frac{1}{2}\) (ĐPCM)

Đặt \(S=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\)

Ta thấy :

\(\frac{1}{201}>\frac{1}{400}\)

\(\frac{1}{202}>\frac{1}{400}\)

...

\(\frac{1}{399}>\frac{1}{400}\)

\(\Rightarrow S>\frac{1}{400}+\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\)

có 200 dãy \(\Rightarrow S>\frac{200}{400}=\frac{1}{2}\)

Vậy : \(S>\frac{1}{2}\)

ai giúp với

Các phân số \(\frac{1}{201};\frac{1}{202};...;\frac{1}{400}\) đều lớn hơn \(\frac{1}{400}\Rightarrow\frac{1}{201}+\frac{1}{202}+...+\frac{1}{400}>\frac{1}{400}.200=\frac{1}{2}\) (do có 200 số hạng)

=> điều phải chứng minh

bn có thể làm cách đầy đủ hơn k Phạm Hồng Quyên

1/201 + 1/202 + ... + 1/400 > 1/400 x 200

1/201 + 1/202 + ... + 1/400 > 1/2

Vậy 1/201 + 1/202 + ... + 1/400 > 1/2

Đặt \(A=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\)

Vì \(\frac{1}{201}>\frac{1}{202}>...>\frac{1}{399}>\frac{1}{400}\)nên :

\(A< \left(\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\right)\)( Có 200 số )

\(A< \frac{1}{400}\times200\)

\(A< \frac{200}{400}\)

\(A< \frac{1}{2}\)( Điều phải chứng minh )

Mk sai chỗ A > ... ( Có 200 số nha )

Chỗ đó phải là A >...

Mong bn thông cảm =))

\(A=16-\frac{\left(-2\right)\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)

\(A=16-\frac{-2}{\frac{1}{3}}=16-\left(-6\right)=22\)

Vậy A = 22 

lấy vở bồi dưỡng toán ra xem ^^ ko có thì thôi^^

câu 1 ; câu 2 tính ra rồi so sánh

giải:

a) A = \(\frac{10}{27}\) +\(\frac{9}{16}\)+\(\frac{11}{34}\)=1,256 

=> A < 2

b) B = \(\frac{1}{12}\)+\(\frac{1}{13}\)+\(\frac{1}{14}\)+....+\(\frac{1}{22}\)=0,477  ; \(\frac{1}{2}\)=0,5

=> ko tke chúng tỏ vì B < 0,5

hok tốt nha !!!

\(a;\)

\(=0-1\)

\(=-1\)

\(b;\)

\(=0-4\)

\(=-4\)

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{15}\)

\(\frac{181\left(x+1\right)}{660}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{181\left(x+1\right)}{660}=\frac{17\left(x+1\right)}{52}\)

\(2353\left(x+1\right)=2805\left(x+1\right)\)

\(2353x+2353=2805x+2805\)

\(2353=2805x+2805-2353x\)

\(2353=452x+2805\)

\(2353-2805=452x\)

\(-452=452x\)

\(x=-1\)

\(S=\frac{\left(9\frac{3}{8}:5,2+3,4.2\frac{7}{34}\right):1\frac{9}{16}}{0,31.8\frac{2}{2}-5,61:27\frac{1}{3}}\)\(\Rightarrow S=\frac{\left(\frac{75}{8}.\frac{5}{26}+\frac{17}{5}.\frac{75}{34}\right):\frac{25}{16}}{\frac{31}{100}.9-\frac{561}{100}.\frac{3}{82}}\)\(\Rightarrow S=\frac{\left(\frac{75.5}{8.26}-\frac{17.75}{5.34}\right).\frac{16}{25}}{\frac{31.9}{100}-\frac{561.3}{100.82}}\)

\(\Rightarrow S=\frac{\left(\frac{375}{208}-\frac{15}{2}\right).\frac{16}{25}}{\frac{279}{100}-\frac{1682}{8200}}\)\(\Rightarrow S=\frac{\frac{-1185}{208}.\frac{16}{25}}{\frac{21196}{8200}}\)\(\Rightarrow S=\frac{-237}{65}:\frac{21196}{8200}\)\(\Rightarrow S=\frac{-194340}{137774}\)

\(\Rightarrow x=\frac{2}{3}S\Rightarrow x=\frac{2}{3}.\frac{-194340}{137774}\Rightarrow x=\frac{-388680}{413322}\)

\(M=\frac{23\frac{11}{15}-26\frac{13}{20}}{12^2+5^2}:\frac{1-\frac{1}{3}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2}-\frac{19}{37}\)\(\Rightarrow M=\frac{\frac{356}{15}-\frac{533}{20}}{12^2+5^2}:\frac{\frac{5}{8}}{3^2.13.2}-\frac{19}{37}\)

\(\Rightarrow M=\frac{\frac{-35}{12}}{12^2+5^2}.\frac{3^2.13.2}{\frac{5}{8}}-\frac{19}{37}\)\(\Rightarrow M=\frac{-84}{13}-\frac{19}{37}\Rightarrow M=\frac{-3355}{481}\Rightarrow15\%M=\frac{-3355}{481}.15\%\Rightarrow15\%M=\frac{-2013}{1924}\)