bằng 2 .Qúa dễ

1 + 1 = 2

Duyệt đi bạn !!!

1+1=2 nhớ 250 điểm nhé!

j mà  nhìu zu zậy làm bao giờ mới xong

Ủng hộ mk đi các bạn
 

Dài quá, chả ai muốn làm, okay?

toi meo 

mik biet lam nhung ngại viet lam ban thong cam nha

\(A=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)\)

Nhận xét:

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}=\frac{1}{150}.50=\frac{1}{3}\)

\(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}=\frac{1}{200}.50=\frac{1}{4}\)

=> A> 1/3 + 1/4 = 7/12 => đpcm

ƠI GIÚP MK CÂU NHƯ THẾ NHƯNG  CMRA>5/8

BN NHA

tích mình đi

làm ơn

rùi mình

tích lại

thanks

\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)

\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)

\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)

\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)

\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)

\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)

\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)

\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)

\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)

cái này dễ lắm chỉ là chưa để ý thôi:

a,1/101>1/102>...>1/199>1/200

=>1/101+1/102+...+1/199+1/200<100*1/101=100/101<1

các phần khác làm tương tự

đánh mỏi tay quá duyệt luôn đi

cái này ở trong học tốt toán 6 đúng không

c) P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

Dễ thấy \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}.50=\dfrac{1}{3}\)(1)

Tương tự

 \(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>50.\dfrac{1}{200}=\dfrac{1}{4}\)(2) 

Từ (1) và (2) ta được

\(P>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) 

P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

         \(\overline{50\text{ hạng tử }}\)                            \(\overline{50\text{ hạng tử }}\)

\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\) 

\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Rightarrow P< \dfrac{5}{6}< 1\)