23/33+2223/3333+222223/333333 > 2 nha 

Vì 23/33+2223/3333+222223/333333 bằng 2.03060606 nha

IV: Để C là số nguyên thì 2n+11⋮n-1

=>2n-2+13⋮n-1

=>13⋮n-1

=>n-1∈{1;-1;13;-13}

=>n∈{2;0;14;-12}

II:

b: \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{59\cdot60}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{59}-\frac{1}{60}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{60}-2\left(\frac12+\frac14+\cdots+\frac{1}{60}\right)\)

\(=1+\frac12+\cdots+\frac{1}{60}-1-\frac12-\cdots-\frac{1}{30}=\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{60}\)

=B

\(\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

=\(\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

=\(\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)

=\(\frac{231}{4}.\frac{4}{21}\)

= 11

\(\dfrac{-7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33.101}{20.101}.\dfrac{33.10101}{30.10101}.\dfrac{33.1010101}{42.1010101}\right)\)

\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33}{20}.\dfrac{33}{30}.\dfrac{33}{42}\right)\)

\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=\dfrac{-7}{4}.33.\dfrac{4}{21}\)

\(=33.\dfrac{-1}{3}\)

\(=11\)

xl nhé kết quả là -11 mk viết thiếu dấu -

\(\frac{2224}{6290}=\frac{1112}{3145}\)

Ta có \(\frac{1112}{3145}< \frac{1112+618}{3145+618}=\frac{1730}{3763}< \frac{2223}{3763}\)

Vậy \(\frac{2223}{3763}>\frac{2224}{6290}\)

Ta có \(\frac{2224}{6290}=\frac{2223}{6290}+\frac{1}{6290}\)

Vì:\(\frac{2223}{3763}>\frac{2223}{6290}\)

Nên\(\frac{2223}{3763}>\frac{2224}{6290}\)

Kết quả :    Viết lại biểu thức đã cho 

           =>         -7/4x . ( 33/12 + 33/20 + 33/30 + 33/42  ) = 22 

                    -7/4x . 33 . ( 1/12 + 1/20 + 1/30 + 1/42 )   = 22 

           -231/4x    . ( 1/3 . 4 + 1/ 4. 5 + 1/5 . 6 + 1/ 6. 7 ) = 22

-231/4x    . ( 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 ) = 22 

                                                -231/4x    . ( 1/3 - 1/7 ) = 22  

                                                -231/4x     .  4/21         = 22 

                                                                -11x            = 22 

                                                                    x             = 22 : -11

                                                                    x             = -2 

                                                      Vậy        x             = -2