khi nao 1 chia 2 bang 1 

\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{98.101}\)

\(=\frac{5}{2}-\frac{5}{5}+\frac{5}{5}-\frac{5}{8}+....+\frac{5}{98}-\frac{5}{101}\)

\(=\frac{5}{2}-\frac{5}{101}=\frac{495}{202}\)

\(\frac{5}{2\times5}+\frac{5}{5\times8}+\frac{5}{8\times11}+...+\frac{5}{98\times101}\)

\(=\frac{5}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{98\times101}\right)\)

\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{5}{3}\times\frac{99}{202}=\frac{165}{202}\)

vi Long Quan rat ban ko co thoi gian ranh

Bài 1.

a) x³ + 2x² - 3x - 6 = ( x³ + 2x² ) - ( 3x + 6 ) = x²( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x² - 3 )

b) ( x - 9 )( x - 7 ) + 1 = x² - 16x + 63 + 1 = x² - 16x + 64 = ( x - 8 )²

c) ( x² + x - 1 )² + 4x² + 4x 

= ( x² + x - 1 )² + 4( x² + x ) (1)

Đặt t = x² + x

(1) <=> ( t - 1 )² + 4t

       = t² - 2t + 1 + 4t

       = t² + 2t + 1

       = ( t + 1 )²

       = ( x² + x + 1 )²

d) ( x² + y² - 17 )² - 4( xy - 4 )²

= ( x² + y² - 17 )² - 2²( xy - 4 )²

= ( x² + y² - 17 )² - [ 2( xy - 4 ) ]²

= ( x² + y² - 17 )² - ( 2xy - 8 )²

= [ ( x² + y² - 17 ) - ( 2xy - 8 ) ][ ( x² + y² - 17 ) + ( 2xy - 8 ) ]

= ( x² + y² - 17 - 2xy + 8 )( x² + y² - 17 + 2xy - 8 )

= [ ( x² - 2xy + y² ) - 17 + 8 ][ ( x² + 2xy + y² ) - 17 - 8 ]

= [ ( x - y )² - 9 ][ ( x + y )² - 25 ]

= [ ( x - y )² - 3² ][ ( x + y )² - 5² ]

= ( x - y - 3 )( x - y + 3 )( x + y - 5 )( x + y + 5 )

Bài 2.

ĐK : x, y ∈ Z

a) x + 2y = xy + 2

<=> x + 2y - xy - 2 = 0

<=> ( x - xy ) - ( 2 - 2y ) = 0

<=> x( 1 - y ) - 2( 1 - y ) = 0

<=> ( 1 - y )( x - 2 ) = 0

+) Nếu 1 - y = 0 => y = 1 và nghiệm đúng với mọi x ∈ Z

+) Nếu x - 2 = 0 => x = 2 và nghiệm đúng với mọi y ∈ Z 

Vậy phương trình có hai nghiệm 

1. \(\hept{\begin{cases}y=1\\\forall x\inℤ\end{cases}}\); 2. \(\hept{\begin{cases}x=2\\\forall y\inℤ\end{cases}}\)

b) xy = x + y

<=> xy - x - y = 0

<=> ( xy - x ) - ( y - 1 ) - 1 = 0

<=> x( y - 1 ) - ( y - 1 ) = 1

<=> ( y - 1 )( x - 1 ) = 1

Ta có bảng sau : 

Các nghiệm trên đều thỏa mãn ĐK

Vậy ( x ; y ) = { ( 2 ; 2 ) , ( 0 ; 0 ) }

a,\(x^3+2x^2-3x-6\)

\(=\left(x^3+2x^2\right)-\left(3x+6\right)\)

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

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

b,\(\left(x-9\right)\left(x-7\right)+1\)

\(=x^2-7x-9x+63+1\)

\(=x^2-16x+64\)

\(=\left(x-8\right)^2\)

1. My father is  interested in playing board games.

2. The last time he smoked cigarettes was a month ago.

3. That apartment is the most modern

4.  Nam's collecttion has over two hundred stamps

5. It takes me 30 minutes to walk to school every morning

6. It's difficult for me to lose weight.

1 My father finds playing board games interesting -> My father is interested in playing board game .
2 He hasn't smoked cigarettes for a month -> The last time he smoked was a month ago .
3 That apartment is more modern than any apartments -> That apartment is the most modern .
4 There are over two hundred stamps in nam's collecition -> Nam's collection has over two hundred stamps.
5 i spend twenty minutes walking to school every morning -> It takes me twenty minutes to walk to school every morning
6 I find losing weight very difficult -> It's difficult for me to losing weight 

a) x( x + 2 )( x + 3 )( x + 5 ) + 5

= [ x( x + 5 ) ][ ( x + 2 )( x + 3 ) ] + 5

= ( x² + 5x )( x² + 5x + 6 ) + 5 (1)

Đặt t = x² + 5x

(1) <=> t( t + 6 ) + 5

       = t² + 6t + 5

       = t² + t + 5t + 5 

       = t( t + 1 ) + 5( t + 1 )

       = ( t + 1 )( t + 5 )

       = ( x² + 5x + 1 )( x² + 5x + 5 )

b) 6x² - 5xy + y² = 6x² - 3xy - 2xy + y² = 3x( 2x - y ) - y( 2x - y ) = ( 2x - y )( 3x - y )

a,\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+5\)

\(=x\left(x+5\right)\left(x+2\right)\left(x+3\right)+5\)

\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+5\)(*)

Đặt \(a=x^2+5x\)ta đc:

(*)=\(a\left(a+6\right)+5\)

\(=a^2+6a+5\)

\(=a^2+a+5a+5\)

\(=a\left(a+1\right)+5\left(a+1\right)\)

\(=\left(a+5\right)\left(a+1\right)\)

\(=\left(x^2+5x+5\right)\left(x^2+5x+1\right)\)

b,\(6x^2-3xy-2xy+y^2\)

\(=3x\left(2x-y\right)-y\left(2x-y\right)\)

\(=\left(3x-y\right)\left(2x-y\right)\)

Ta có : \(\frac{2010}{2011}=\frac{2011}{2011}-\frac{1}{2011}=1-\frac{1}{2011}\)

             \(\frac{2011}{2012}=\frac{2012}{2012}-\frac{1}{2012}=1-\frac{1}{2012}\)

mà \(\frac{1}{2011}>\frac{1}{2012}\)

\(\Rightarrow1-\frac{1}{2011}< 1-\frac{1}{2012}\)

\(\Rightarrow\frac{2010}{2011}< \frac{2011}{2012}\)