Ta có:
\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2016^2}{2015.2017}\)
\(A=\dfrac{1}{2}.\dfrac{2^2}{3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2016^2}{2015.2017}\)
\(A=\left(\dfrac{2.3.4...2016}{2.3.4.5...2015}\right).\left(\dfrac{2.3.4...2016}{2.3.4.5...2017}\right)\)
\(A=2016.\dfrac{1}{2017}=\dfrac{2016}{2017}\)

Theo đề bài ta có: \(\dfrac{52367-n}{47633+n}=\dfrac{17}{83}\)
\(\Leftrightarrow83\left(52367-n\right)=17\left(47633+n\right)\)
\(\Leftrightarrow4346461-83n=809761+17n\)
\(\Leftrightarrow4346461-809761=83n+17n\)
\(\Leftrightarrow3536700=100n\)
\(\Leftrightarrow n=35367\)

Phân tích đa thức thành nhân tử à bạn ?!

Pt \(\Leftrightarrow\dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{5}{12}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{5}{12}\)
\(\Leftrightarrow-\dfrac{1}{6}.x=\dfrac{5}{12}\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)

• Nếu x=0 ta có: \(2^0+624=5^y\)
\(\Leftrightarrow5^y=625=5^4\)
=> y=4
• Nếu x>0 ta thấy VT của pt là số chẵn, VP là số lẻ
=> Pt vô nghiệm
Vậy có 1 cặp số duy nhất thỏa mãn đề bài là (x;y) = (0;4)

a, \(\left|5x-4\right|\ge6\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4\ge6\\5x-4\le-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-\dfrac{2}{5}\end{matrix}\right.\)

Henry plays the guitar wonderful.

Ta có \(\left(\dfrac{11}{23}+\dfrac{31}{33}-\dfrac{49}{33}\right)^0=1\)
=> \(\dfrac{1}{4}+2\dfrac{3}{4}\left(\dfrac{11}{23}+\dfrac{31}{33}-\dfrac{49}{43}\right)^0=\dfrac{1}{4}+\dfrac{11}{4}=\dfrac{12}{4}=3\)
Sao bạn không dùng máy tính ??

It’s necessary to protect our forests from fire.

Từ gt ta có:
\(\dfrac{13}{3}.\left(-\dfrac{1}{3}\right)\le x\le\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\)
\(\Leftrightarrow\dfrac{-13}{9}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow\dfrac{-26}{18}\le x\le-\dfrac{11}{18}\)
Suy ra \(26\ge x\ge11\)
Vậy \(11\le x\le26\) ( x thuộc Z ) là các giá trị cần tìm