\(B=\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(B=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)

\(A=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{49.50}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=-\left(1-\dfrac{1}{50}\right)\)
\(=-\dfrac{49}{50}\)

nhanh thế

ĐKXĐ: x<>5

Để B là số nguyên thì \(2x-11⋮x-5\)

=>\(2x-10-1⋮x-5\)

=>\(-1⋮x-5\)

=>\(x-5\in\left\{1;-1\right\}\)

=>\(x\in\left\{6;4\right\}\)

Chưa đủ dữ liệu em nhé.

đề bài dou bạn?

tính tổng dãy số đó 

trọng lượng của túi hàng là:

\(P=10\cdot34=340\left(N\right)\)

Trọng lượng của An là \(P=10\cdot38=380\left(N\right)\)

Trọng lượng của bao xi măng là \(P=10\cdot50=500\left(N\right)\)

\(\dfrac{2}{-5}-\left(\dfrac{5}{2}-\dfrac{12}{5}\right)\)

\(=-\dfrac{2}{5}-\dfrac{5}{2}+\dfrac{12}{5}=2-\dfrac{5}{2}=-\dfrac{1}{2}\)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)

\(\Rightarrow3\cdot\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)

\(\Rightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}:3\)

\(\Rightarrow \left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{27}\)

\(\Rightarrow\left(3x-\dfrac{1}{2}\right)^3=\left(-\dfrac{1}{3}\right)^3\)

\(\Rightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Rightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}\)

\(\Rightarrow3x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}:3=\dfrac{1}{18}\)

\(-\dfrac{3}{17}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{17}-33\cdot\dfrac{3}{17}\)

\(=-\dfrac{3}{17}\left(\dfrac{5}{9}+\dfrac{4}{9}+33\right)\)

\(=-\dfrac{3}{17}\cdot34=-6\)

-6 ạ

\(\dfrac{26-x}{995}+\dfrac{22-x}{997}+\dfrac{18-x}{999}=\dfrac{12-x}{334}\)

=>\(\dfrac{x-26}{995}+\dfrac{x-22}{997}+\dfrac{x-18}{999}=\dfrac{x-12}{334}\)

=>\(\left(\dfrac{x-26}{995}-2\right)+\left(\dfrac{x-22}{997}-2\right)+\left(\dfrac{x-18}{999}-2\right)=\dfrac{x-12}{334}-6\)

=>\(\dfrac{x-2016}{995}+\dfrac{x-2016}{997}+\dfrac{x-2016}{999}-\dfrac{x-2016}{334}=0\)

=>x-2016=0

=>x=2016