A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)

= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)

= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)

= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)

= \(\dfrac{215}{1}=215\)

B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)

= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)

= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)

= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)

= \(\dfrac{300}{2}=150\)

C = \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{333333}{424242}\right)\)

= \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{11}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

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

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

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

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

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

= \(-\dfrac{7}{4}.\dfrac{4}{21}.33=-\dfrac{1}{3}.33=-11\)

a)

Ta thấy:

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

\(\dfrac{1}{7}< \dfrac{1}{5}\)

\(\dfrac{1}{8}< \dfrac{1}{5}\)

\(\dfrac{1}{9}< \dfrac{1}{5}\)

\(\dfrac{1}{11}< \dfrac{1}{10}\)

\(\dfrac{1}{12}< \dfrac{1}{10}\)

\(\dfrac{1}{13}< \dfrac{1}{10}\)

...

\(\dfrac{1}{17}< \dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 5\cdot\dfrac{1}{5}+8\cdot\dfrac{1}{10}=1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)

Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)

b)

Ta thấy:

\(\dfrac{1}{101}>\dfrac{1}{300}\)

\(\dfrac{1}{102}>\dfrac{1}{300}\)

\(\dfrac{1}{103}>\dfrac{1}{300}\)

...

\(\dfrac{1}{299}>\dfrac{1}{300}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>200\cdot\dfrac{1}{300}=\dfrac{2}{3}\)

Vậy \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)

Bài 2: 

b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ\left(1\right)\)

\(\Leftrightarrow d\in\left\{1;-1\right\}\)

\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)

hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)

Bài 1: 

a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)

\(=75\cdot\left(-4\right)+603\)

\(=603-300=303\)

Bài 1: 

c) Ta có: \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Leftrightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Leftrightarrow3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow2B=1-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow B=\dfrac{3^{99}-1}{3^{99}\cdot2}\)

Bài 2: 

a) Vì tổng của hai số là 601 nên trong đó sẽ có 1 số chẵn, 1 số lẻ

mà số nguyên tố chẵn duy nhất là 2

nên số lẻ còn lại là 599(thỏa ĐK)

Vậy: Hai số nguyên tố cần tìm là 2 và 599

b,Gọi ƯCLN(21n+4,14n+3)=d

21n+4⋮d ⇒42n+8⋮d

14n+3⋮d ⇒42n+9⋮d

(42n+9)-(42n+8)⋮d

1⋮d ⇒ƯCLN(21n+4,14n+3)=1

Vậy phân số 21n+4/14n+3 là phân số tối giản

c,xy-2x+5y-12=0

xy-2x+5y-12+2=0+2

xy-2x+5y-10=2

xy-2x+5y-5.2=-2

x.(y-2)+5.(y-2)=2

(y-2).(x+5)=2

Sau đó bạn tự lập bảng 

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)

\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)

\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)

\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)

\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)

Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)

...

\(\dfrac{1}{299}>\dfrac{1}{300}\)

Do đó :

\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)

Vậy...

A=3/2².8/3².15/4²......899/30² A=3.8.15.....899/2².3².4².....30² A=(1.3).(2.4).(3.5).....(29.31)/(2.3.4....30)(2.3.4...30) A=(1.2.3....29).(3.4.5...31)/(2.3.4...30)(2.3.4...30) A=1.31/30.2=31/60

a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)

\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)

\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)

\(=\dfrac{10}{12}\)

\(=\dfrac{5}{6}\)

\(---\)

b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)

\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)

\(=18+\dfrac{3015}{2}\)

\(=\dfrac{36}{2}+\dfrac{3015}{2}\)

\(=\dfrac{3051}{2}\)

\(---\)

c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)

\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(---\)

d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)

\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)

\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)

\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)

\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)

\(=-\dfrac{25}{41}\)

#\(Toru\)

a,a+1/4=2 3/4-1 1/2    

a+1/2=5/4

    a=5/4-1/2

     a=3/4

b,a-7/4=13/4-7/9

a-7/4=89/36

        a= 89/36+7/4

         a=152/36

c,3/2-a=17/6-1/6

3/2-a=8/3

       a= 3/2-8/3

       a= -7/6

a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)

\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)

\(3-a=\dfrac{3}{2}\)

\(a=3-\dfrac{3}{2}\)

\(a=\dfrac{6}{2}-\dfrac{3}{2}\)

\(a=\dfrac{3}{2}\)

b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)

\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)

\(0-a=\dfrac{7}{8}\)

\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)

c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)

\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)

\(\dfrac{8}{6}-a=\dfrac{1}{6}\)

\(a=\dfrac{8}{6}-\dfrac{1}{6}\)

\(a=\dfrac{7}{6}\)

a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)

     a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)

     a  = 2 + 1 - 1 - \(\dfrac{1}{2}\)

     a  = 2 - \(\dfrac{1}{2}\)

     a = \(\dfrac{3}{2}\)

b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)

    (3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)

                     a = - \(\dfrac{7}{8}\)

c,    2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a  = \(\dfrac{1}{6}\)

    a =  2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\)  - \(\dfrac{1}{6}\) 

     a =  (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)

     a = 1 +  \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)

     a = \(\dfrac{7}{6}\)

`#040911`

`a)`

\(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\\ \left(2\dfrac{3}{4}+\dfrac{1}{4}\right)-a=1\dfrac{1}{2}\\ 3-a=1\dfrac{1}{2}\\ a=3-1\dfrac{1}{2}\\ a=\dfrac{3}{2}\\ \text{Vậy, a = }\dfrac{3}{2}\)

`b)`

\(3\dfrac{1}{4}-a-3\dfrac{1}{4}=\dfrac{7}{8}\\ \left(3\dfrac{1}{4}-3\dfrac{1}{4}\right)-a=\dfrac{7}{8}\\0-a=\dfrac{7}{8}\\ a=0-\dfrac{7}{8} \\ a=\dfrac{-7}{8}\)

Bạn xem lại đề, lớp 5 chưa học dấu âm.

`c)`

\(2\dfrac{5}{6}-1\dfrac{1}{2}-a=\dfrac{1}{6}\\ \dfrac{4}{3}-a=\dfrac{1}{6}\\ a=\dfrac{4}{3}-\dfrac{1}{6}\\ a=\dfrac{7}{6}\\ \text{Vậy, a = }\dfrac{7}{6}.\)

b\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4

Tương tự như vậy với câu a\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2