\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

Có \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\)

nhan xet:3/1.4=1/1-1/4

3/4.7=1/4-1/7

3/7.10=1/7-1/10

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

3/40.43=1/40-1/43

3/43.46=1/43-1/46

S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S=1/1-1/46

S=46/46-1/46

S=45/46<1

vay s<1

S= 1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+|1/43-1/46

S= 1-1/46

S= 45/46<1

vậy S<1

duyệt đi

S=  1- 1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S= 1+ (1/4-1/4)+(1/7-1/7)+...+(1/43-1/43)-1/46

S= 1-1/46= 45/46<1

Suy ra S<1

S=1 - 1/46 < 1 

Có \(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{43.46}\)

Sẽ có: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{43.46}\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+........+\frac{1}{43}-\frac{1}{46}\)

\(S=\frac{1}{1}-\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+....+\left(-\frac{1}{43}+\frac{1}{43}\right)-\frac{1}{46}\)

\(S=\frac{1}{1}-0+0+0+0+......+0+0-\frac{1}{46}\)

\(S=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)

Vì có:  \(\frac{45}{46}<1\) nên \(S<1\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

\(S=\frac{45}{46}\)

=> \(\frac{45}{46}<\frac{46}{46}=1\)

=> S<1

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

ta có 

\(\frac{3}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{1}{4}-\frac{1}{7}\)

.....

\(\frac{3}{43.46}=\frac{1}{43}-\frac{1}{46}\)

( mình nghĩ cậu chưa đc làm dạng như này nên ghi ra , lần sau có gặp mà biết cách làm  r thì bỏ bước trên đi cx đc nha)

\(=>S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=>S=1-\frac{1}{46}< 1\)(dpcm

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

    \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

      \(=1-\frac{1}{46}< 1\)

Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)

=>S= 1- 1/4 + 1/4 -1/7 + 1/7 - 1/10 +...+ 1/n - 1/(n+3)

=>S= 1- 1/(n+3)

=>S + 1/(n+3) = 1

=>S<1

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

A = \(\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

A = 1 + \(\frac{1}{4}\) - \(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) -....- \(\frac{1}{2011}\) + \(\frac{1}{2014}\)

A = 1 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\Rightarrow\frac{99}{100}=\frac{0.33.x}{2009}\)

\(\Rightarrow100.0.33.x=99.2009\)

\(\Rightarrow0x=198891\Rightarrow\)không có GT x thỏa mãn