`#3107`

`a)`

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=1-\dfrac{1}{2000}\)

\(=\dfrac{1999}{2000}\)

`b)`

\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

`c)`

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

b) Sửa đề:

 \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

\(-\dfrac{3}{4}.31\dfrac{11}{23}-0,75.8\dfrac{12}{23}\)

\(=-\dfrac{3}{4}.31\dfrac{11}{23}-\dfrac{3}{4}.8\dfrac{12}{23}\)

\(=-\dfrac{3}{4}.\left(31+\dfrac{11}{23}+8+\dfrac{12}{12}\right)\)

\(=-\dfrac{3}{4}.\left(31+8+1\right)\)

\(=-\dfrac{3}{4}.40\)

\(=-3.10\)

\(=-30\)

- \(\dfrac{3}{4}\).31\(\dfrac{11}{23}\) - 0,75.8\(\dfrac{12}{23}\)

= - \(\dfrac{3}{4}\).\(\dfrac{724}{23}\) - \(\dfrac{3}{4}\)\(\dfrac{196}{23}\)

=   - \(\dfrac{3}{4.23}.\left(724+196\right)\)

= - \(\dfrac{3}{92}\) . 920

= - 30 

`(3/4)^5 * x=(3/4)^7`

`=> x= (3/4)^7 : (3/4)^5`

`=> x= (3/4)^(7-5)`

`=>x=(3/4)^2`

`=>x= 9/16`

\(\left(\dfrac{3}{4}\right)^5\cdot x=\left(\dfrac{3}{4}\right)^7\)

\(x=\left(\dfrac{3}{4}\right)^7\div\left(\dfrac{3}{4}\right)^5=\left(\dfrac{3}{4}\right)^2\)

Gọi diện tích mỗi thửa ruộng theo thứ tự từ bé đến lớn lần lượt là: \(x\); y (đk : \(x\); y > 0)

Theo bài ra ta có: \(\dfrac{x}{3.3}\) = \(\dfrac{y}{4.4}\)  = \(\dfrac{x}{9}\) = \(\dfrac{y}{16}\) 

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

                             \(\dfrac{x}{9}\)  = \(\dfrac{y}{16}\) = \(\dfrac{x+y}{9+16}\) = \(\dfrac{10000}{25}\) = 400

⇒ \(x\) = 400 x 9 = 3 600 (m²)

    y = 400 x 16 = 6 400 (m²)

Kết luận .......

Ta thấy : \(x^2\ge0=>x^2+2\ge2>0\forall x\)

Do đó phương trình trên vô nghiệm 

Ta có: \(A=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=\dfrac{x-2}{x-2}+\dfrac{3}{x-2}=1+\dfrac{3}{x-2}\)

Để A là số nguyên thì \(x-2\inƯ\left(3\right)=\left\{-1,-3,1,3\right\}\)

Ta có bảng giá trị:

Vậy ...

Ta có : \(A=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}\)

\(\Rightarrow A=1+\dfrac{3}{x-2}\)

Vì x là số nguyên nên để A cũng là số nguyên thì : \(\dfrac{3}{x-2}\in Z\)

\(\Rightarrow3⋮\left(x-2\right)\)

\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)\)

Do đó ta có bảng :

Vậy..........