\(I=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2009.2010}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}\\ =\dfrac{2009}{2010}\)

\(K=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{2008.2010}\\ =2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\\ =2.\dfrac{502}{1005}\\ =\dfrac{1004}{1005}\)

\(F=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}.\dfrac{10}{33}\\ =\dfrac{10}{99}\)

\(I=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}=\dfrac{2010-1}{2010}=\dfrac{2008}{2010}\)

\(K=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}=\dfrac{502}{1005}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{99.100}=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{100}\right)=\dfrac{97}{900}\)

\(y:\dfrac{3}{11}=\dfrac{7}{6}+\dfrac{2}{3}\)

\(y:\dfrac{3}{11}=\dfrac{11}{6}\)

\(y=\dfrac{11}{6}\times\dfrac{3}{11}\)

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

Bài 3

\(\dfrac{3}{7}:\dfrac{1}{4}-\dfrac{1}{2}=\dfrac{3}{7}\times\dfrac{4}{1}-\dfrac{1}{2}=\dfrac{12}{7}-\dfrac{1}{2}=\dfrac{24}{14}-\dfrac{7}{14}=\dfrac{17}{14}\)

`2)`

`y:3/11=7/6+2/3`

`y:3/11=11/6`

`y=11/6xx3/11`

`y=1/2`

`3)`

`=12/7-1/2=24/14-7/14=17/14`

`7/12 + 3/4 xx 2/9`

`= 7/12 + 1/6`

`= 9/12`

_________________

`8/9 - 4/15 : 2/5`

`= 8/9 - 2/3`

`= 2/9`

giúp me 

a) _{^{\(\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{3}{4}\)}}

b) _{^{\(\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)}}

\(A=\dfrac{31\cdot\left(31^{12}-1\right)}{31\left(31^{13}+1\right)}=\dfrac{31^{13}+1-32}{31\left(31^{13}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{14}+31}\)

\(B=\dfrac{31\left(31^{13}-1\right)}{31\left(31^{14}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)

Dễ thấy \(31^{14}+31< 31^{15}+31\Rightarrow\dfrac{32}{31^{14}+31}>\dfrac{32}{31^{15}+31}\\ \Rightarrow\dfrac{1}{31}-\dfrac{32}{31^{14}+31}< \dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)

Vậy A < B

c) \(=\left(4x-3\right)^2-\left(9x^2-4\right)\)

\(=16x^2-24x+9-9x^2+4=7x^2-24x+13\)

d) \(=\left(x^2-3x+2\right)\left(x+3\right)-\left(x^3-5x^2\right)\)

\(=x^3+3x^2-3x^2-9x+2x+6-x^3+5x^2\)

\(=5x^2-7x+6\)

c. (4x - 3)(4x - 3) - (3x + 2)(3x - 2)

= (4x - 3)² - (9x² - 4)

= 16x² - 24x + 9 - 9x² + 4

= 16x² - 9x² - 24x + 9 + 4

= 7x² - 24x + 13

d. (x - 2)(x - 1)(x + 3) - x²(x - 5)

= (x² - 1 - 2x + 2)(x + 3) - x²(x - 5)

= x³ + 3x² - x - 3 - 2x² - 6x + 2x + 6 - x³ + 5

= x³ - x³ + 3x² - 2x² - x - 6x + 2x + 6 + 5 - 3

= x² - 5x + 8

thi tự làm nhé bạn

Câu 1:

\(a,\text{Slt: }\widehat{A_4}\text{ và }\widehat{B_1}\\ \text{Đv: }\widehat{A_4}\text{ và }\widehat{B_3}\\ b,a\text{//}b\Rightarrow\left\{{}\begin{matrix}\widehat{A_4}=\widehat{B_1}=42^0\left(\text{so le trong}\right)\\\widehat{B_2}=180^0-\widehat{A_4}=138^0\left(\text{trong cùng phía}\right)\end{matrix}\right.\)

P=.....xem thêm

= 26 x 182 + 182 x 47 + 91 x 27 x 2

=  26 x 182 + 182 x 47 + 182 x 27

= 182 x ( 26 + 47 + 27)

= 182 x 100 = 18200

 26 x 182 + 182 x  47 + 91 x 27 x 2

= 26 x 182 + 182 x  47 + ( 91 x 2 ) x 27

= 26 x 182 + 182 x  47 + 182 x 27

= 182 x ( 26 + 27 + 47 )

= 182 x 100

= 18200

= 26 x 182 + 182 x 47 + 91 x 27 x 2

= 26 x 182 + 182 x 47 + 182 x 2 x 27

= 182 x (26 + 47 + 27)

= 182 x 100 

= 18200

1 A

2 C

3 B

4 D

5 A

6 C

7 D

8 D

1 A

2 C

3 B

4 D

5 A