1 - 2 + 3 - 4 + ............ + 4019 - 4020 (4020 số)

= (1 - 2) + (3 - 4) + ......... + (4019 - 4020)   [2010 cặp]

= (-1) + (-1) + ............ + (-1)

= (-1) . 2010 

= -2010

=(-1)+(-1)+(-1)+.....+(-1)   (có 2010 số nguyên -1)   =(-1)*2010  =-2010

Đặt S = 1 - 2 + 3 - 4 + 5 - 6 + ........... + 4019 - 4020

      S = ( 1 - 2 ) + ( 3 - 4 ) + ( 5 - 6 ) + ......... + ( 4019 - 4020 )

      S =   ( -1 )   +   ( -1 )   +   ( -1 )   + .......... +    ( -1 )

                        ( có 2010 số hạng ( -1 ) )

      S = ( -1 ) . 2010

      S =  -2010
 

Chọn D

Chọn D

\(\left(\frac{4}{2.4}+\frac{4}{4.6}+.....+\frac{4}{4020.4022}\right)x=2010\)

\(\Leftrightarrow2x\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{4020.4022}\right)=2010\)

\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4}+.....+\frac{1}{4020}-\frac{1}{4022}\right)=2010\)

\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4022}\right)=2010\)

Tự biên tự diễn

Ko chép lại đề nhé

<=> 2( 2/2.4 + 2/2.6 + 2/2.8 +...+ 2/ 4020.4022) x= 2010

<=> 2( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +....+ 1/4020- 1/4022 )x=2010

<=> ( 1/2 - 1/4022)2x = 2010

<=> ( 2011/4022 - 1/4022 )2x = 2010

<=>( 2010/4022) .2x= 2010

<=> 2x = 2010 : 2010/4022

<=> 2x = 4022

=> x = 2011 

Vậy x = 2011

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x.\left(x+1\right)}=1\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x.\left(x+1\right)}=1\)

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

MK chưa làm xong đợi mk ăn cơm xong làm nha

Chắc câu a là \(2a^2\) ...

\(A=\left(a^2+b^2+2ab-4a-4b+4\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+4014\)

\(A=\left(a+b-2\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+4014\ge4014\)

\(A_{min}=4014\) khi \(a=b=1\)

\(B=\left(x^2-7x\right)\left(x^2-7x+12\right)=\left(x^2-7x\right)^2+12\left(x^2-7x\right)\)

\(B=\left(x^2-7x+6\right)^2-36\ge-36\)

\(B_{min}=-36\) khi \(\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)

C

C

Chọn C