j mà  nhìu zu zậy làm bao giờ mới xong

Ủng hộ mk đi các bạn
 

câu b 

C= 1/181+1/182+...1/200< 20/200=1/10 
A=B+C<4/9+1/10=40/90+9/90=49/90 mà 49/90<3/4 ( quy đồng)
Vậy A<3/4 
** D= 1/101+1/101+...1/150>50.(1/101)=50/101>1... 
E= 1/151+1/152+...+1/200> 50.(1/151)=50/151>1/3 
D+E>1/3+1/3=2/3 mà 2/3>5/8 
Vậy A>5/8

a)Ta CM: S(n)>7/12 (*) bằng qui nạp 
+S(3)=1/4+1/5+1/6>7/12 
+giã sử S(k)>7/12 (k>=3, k nguyên) 
tức là:S(k)=1/(k+1)+1/(k+2)+...+1/2k>7/12 
+Ta có: S(k+1)=1/(k+2)+1/(k+3)+...+1/(2k+2) 
=1/(k+1)+1/(k+2)+... 
..+1/2k+1/(2k+1)+1/(2k+2)-1/(k+1) 
=S(k)+1/(2k+1)+1/(2k+2)-1/(k+1) 
=S(k)+1/[(2k+1)(2k+2)]>7/2 
theo nguyên lí qui nạp=>(*) đúng với mọi n>3, n nguyên

câu b tương tự

Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)

\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)

\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)

\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)

hay \(C>\dfrac{5}{8}\)(đpcm)

Ta có: 𝐶=1101+1102+1103+...+1200C=1011​+1021​+1031​+...+2001​

=(1101+1102+...+1120)+(1121+1122+1123+...+1150)+(1151+1152+1153+...+1180)+(1181+1182+1183+...+1200)=(1011​+1021​+...+1201​)+(1211​+1221​+1231​+...+1501​)+(1511​+1521​+1531​+...+1801​)+(1811​+1821​+1831​+...+2001​)

⇔𝐶>20⋅1120+30⋅1150+30⋅1180+20⋅1200⇔C>20⋅1201​+30⋅1501​+30⋅1801​+20⋅2001​

⇔𝐶>16+15+16+110=1930=76120⇔C>61​+5

Ta có:

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

\(c=\)(\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\)...\(+\)\(\frac{1}{120}\))\(+\)(\(\frac{1}{121}\)\(+\)\(\frac{1}{122}\)\(+\)...\(+\)\(\frac{1}{150}\))\(+\)(\(\frac{1}{151}\)\(+\)\(\frac{1}{152}\)\(+\)...\(+\)\(\frac{1}{180}\))\(+\)(\(\frac{1}{181}\)\(+\)\(\frac{1}{182}\)\(+\)...\(+\)\(\frac{1}{200}\))>20\(.\)\(\frac{1}{120}\)\(+\)30\(.\)\(\frac{1}{150}\)\(+\)30\(.\)\(\frac{1}{180}\)\(+\)20\(.\)\(\frac{1}{200}\)= \(\frac{1}{6}+\frac{1}{5}\)\(+\)\(\frac{2}{6}+\frac{1}{10}\)= \(\frac{19}{30}\)=\(\frac{76}{120}\)> \(\frac{75}{120}\)=\(\frac{5}{8}\)

=>\(c\)>\(\frac{5}{8}\)(đpcm)

_Hok tốt_

Cảm ơn nhiều !