A

Chọn A

A

B = ( -1 ) + 2 + ( -3 ) + 4 + ... + ( -49 ) + 50

=> B = [ 2 + ( -1 ) ] + [ ( -3 ) + 4 ] + ... + [ ( -49 ) + 50 ] ( 25 cặp số )

=> B = 1 + 1 + ... + 1

=> B = 1 x 25

=> B = 25

Vậy B = 25

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)

\(=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+\frac{1}{\frac{4.\left(4+1\right)}{2}}+.....+\frac{1}{\frac{50\left(50+1\right)}{2}}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)

\(C=\left(1-2-3-4\right)+...+\left(197-198-199-200\right)\)

=-8x25=-200

\(D=-\left(11+13+...+99\right)+\left(10+12+...+98\right)\)

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

=-1x45=-45

Ta có:

\(x^2-2018x+1=0\)

\(\Leftrightarrow x^2+1=2018x\)

Do đó

\(B=\frac{x^4+x^2+1}{x^2}=\frac{\left(x^4+2x^2+1\right)-x^2}{x^2}=\frac{\left(x^2+1\right)^2-x^2}{x^2}=\frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2}\)

\(\Leftrightarrow B=\frac{\left(2018x+x\right)\left(2018x-x\right)}{x^2}=\frac{2019x\cdot2017x}{x^2}=2019\cdot2017\)

A=-1275