nhanh giúp mình

Bài 1:

1: \(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{11\cdot13}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{11\cdot13}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac12\left(\frac13-\frac{1}{13}\right)=\frac12\cdot\frac{10}{39}=\frac{5}{39}\)

2: \(C=\frac12+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(=\frac24+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}+\frac{2}{304}\)

\(=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}+\frac{2}{16\cdot19}\)

\(=\frac23\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\right)\)

\(=\frac23\left(1-\frac14+\frac14-\frac17+\frac17-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac23\left(1-\frac{1}{19}\right)=\frac23\cdot\frac{18}{19}=\frac{2\cdot6}{19}=\frac{12}{19}\)

Bài 2:

\(\frac{1}{101}>\frac{1}{300};\frac{1}{102}>\frac{1}{300};\ldots;\frac{1}{299}>\frac{1}{300};\frac{1}{300}=\frac{1}{300}\)

Do đó: \(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+\cdots+\frac{1}{300}=\frac{200}{300}=\frac23\) (ĐPCM)

Câu hỏi của Hoàng Đỗ Việt - Toán lớp 6 | Học trực tuyến

Bài 1 :

Ta có;\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}>\frac{1}{30}.10=\frac{1}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}.30>\frac{1}{30}.24=\frac{2}{5}\)

Do đó :

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\left(1\right)\)

Mặt khác :

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}.20=1\)

\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}.20=\frac{1}{2}\)

Do đó :

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< 1+\frac{1}{2}=\frac{3}{2}\left(2\right)\)

Từ (1 ) và (2) ta suy ra điều phải chứng minh

Bài 2 : 

Đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)

MỘT MẶT ,TA CÓ THỂ VIẾT

\(S=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)\(+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}\right)-\frac{1}{64}\)

\(>\frac{1}{2}.2+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32-\frac{1}{64}\)\(=\frac{7}{2}-\frac{1}{64}=\frac{223}{64}>\frac{192}{64}=3\left(1\right)\)

Mặt khác ,ta lại có\(S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\)\(+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)\)\(+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)< \)\(1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32=6\left(2\right)\)

Từ (1) và (2 ) ta kết luận \(3< S< 6\)

Chúc bạn học tốt ( -_- )

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

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

a) CÓ: A = (1-1/4²).(1-1/5²).(1-1/6²)......(1-1/200²)

               =\(\frac{4^2-1^2}{4^2}\). \(\frac{5^2-1^2}{5^2}\). \(\frac{6^2-1^2}{6^2}\)....... \(\frac{200^2-1^2}{200^2}\)

Ta có công thức sau : a²-b²= a² -ab+ab-b² 

                                            = a(a-b) + b(a-b)

                                            = (a+b)(a-b)

   ÁP DỤNG CÔNG THỨC TRÊN VÀO BÀI TOÁN TA ĐƯỢC : 

  A=  \(\frac{3.5}{4^2}\). \(\frac{4.6}{5^2}\). \(\frac{5.7}{6^2}\)......\(\frac{199.201}{200^2}\)

    = \(\frac{\left(3.4.5.....199\right)\left(5.6.7....201\right)}{\left(4.5.6......200\right)^2}\)

    =    \(\frac{\left(3.4.5.......199\right)\left(5.6.7.....200.201\right)}{\left(4.5.6.....199.200\right)\left(4.5.6......200\right)}\)

    =   \(\frac{3.201}{200.4}\)

   =  \(\frac{603}{800}\)​​

b)Từ đề bài ta suy ra : B=\(\frac{1.3}{5.7}\).\(\frac{3.5}{7.9}\). \(\frac{5.7}{9.11}\)...... \(\frac{99.101}{103.105}\)

                                      = \(\frac{1.3^2.5^2.7^2......99^2.101}{5.7^2.9^2.11^2....99^2.101^2.103^2.105}\)

                                      =\(\frac{3^2.5}{101.103^2.105}\)

                                       =\(\frac{3}{7500563}\)

2/\(\frac{10}{11}\)

câu 1:

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)

\(\hept{\begin{cases}15=3\cdot5\\35=5\cdot7\end{cases}}\\ \hept{\begin{cases}3=3\\63=3^2\cdot7\\99=3^2\cdot11\end{cases}}\)

=>\(\frac{2310}{3465}+\frac{462}{3465}+\frac{198}{3465}+\frac{110}{3465}+\frac{70}{3465}\)

=>\(\frac{2310+462+198+110+70}{3465}\)

=>\(\frac{3150}{3465}\)=\(\frac{10}{11}\)