\(=\left[\dfrac{9-3}{162}\cdot\dfrac{81}{17}+\dfrac{35}{34}\right]:\left(\dfrac{3}{5}+\dfrac{7}{102}\right)\cdot\dfrac{102}{5}+2017\)

\(=\left[\dfrac{6}{2}\cdot\dfrac{1}{17}+\dfrac{35}{34}\right]:\dfrac{341}{510}\cdot\dfrac{102}{5}+2017\)

\(=\dfrac{41}{34}\cdot\dfrac{510}{341}\cdot\dfrac{102}{5}+2017=\dfrac{12546}{341}+2017=\dfrac{700343}{341}\)

Ta có :\(A=\left(5+10+15+...+1000\right).\left\{\frac{2}{5}:0,5+2:\left(-0,4\right)\right\}:\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1000}\right)\)

\(\Leftrightarrow A=\left(5+10+15+...+1000\right).\left\{\frac{2}{5}:\frac{1}{2}+2:\left(-\frac{2}{5}\right)\right\}:\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1000}\right)\)\(\Leftrightarrow A=\left(5+10+15+...+1000\right).\left\{\frac{2}{5}.2+2.\left(-\frac{5}{2}\right)\right\}:\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1000}\right)\)\(\Leftrightarrow A=\left(5+10+...+1000\right).\left\{2.\left(\frac{2}{5}-\frac{5}{2}\right)\right\}.\left(5+10+...+1000\right)\)

\(\Leftrightarrow A=\left(5+10+...+1000\right).\left(5+10+...+1000\right).-\frac{21}{10}\)

Ta có : Số số hạng của dãy số : \(5+10+...+1000\) là :

\(\left(1000-5\right):5+1=200\)

\(\Rightarrow\) Tổng của dãy số : \(5+10+...+1000\) là :

\(\frac{\left(5+1000\right).200}{2}=100500\)

\(\Rightarrow A=100500.100500.\left(-\frac{21}{10}\right)\)

\(\Rightarrow A=100500^2.\left(-\frac{21}{10}\right)\)

\(\Rightarrow A=\frac{100500^2.\left(-21\right)}{10}\)

Vậy :\(A=\frac{100500^2.\left(-21\right)}{10}\)

P/s: Số to quá nên mình đề dưới dạng phân số, không tính ra kết quả cụ thể.

bài này dễ quá nên mik không làm

Dễ thì làm đi

nghĩ cũng sẽ ra = 999/100 nhé đúng thì nhấn nhé ủng hộ nha 

\(1,\\ c,\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\dfrac{17\times7-3\times6-5\times7}{6\times7}=\dfrac{119-18-35}{42}=\dfrac{66}{42}=\dfrac{11}{7}\\ 2,\\ a,\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\\ =\dfrac{6:3}{9:3}+\dfrac{7+3}{15}\\ =\dfrac{2}{3}+\dfrac{10}{15}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{2+2}{3}=\dfrac{4}{3}\)

\(b,\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\\ =\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\\ =\dfrac{7}{2}+\dfrac{29-11}{12}\\ =\dfrac{7}{2}+\dfrac{18}{12}\\ =\dfrac{7}{2}+\dfrac{18:6}{12:6}\\ =\dfrac{7}{2}+\dfrac{3}{2}\\ =\dfrac{7+3}{2}=\dfrac{10}{2}=5\)

\(c,\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\\ =\dfrac{26}{25}-\dfrac{3\times5}{5\times5}-\dfrac{2\times5}{5\times5}\\ =\dfrac{26-15-10}{25}\\ =\dfrac{1}{25}\)

Bài 1:
\(\dfrac{17}{6}-\dfrac{3}{7}-\dfrac{5}{6}=\left(\dfrac{17}{6}-\dfrac{5}{6}\right)-\dfrac{3}{7}=2-\dfrac{3}{7}=\dfrac{14}{7}-\dfrac{3}{7}=\dfrac{11}{7}\)
Bài 2:
a/\(\dfrac{6}{9}+\dfrac{3}{15}+\dfrac{7}{15}\)
\(=\dfrac{2}{3}+\left(\dfrac{3}{15}+\dfrac{7}{15}\right)\)
\(=\dfrac{2}{3}+\dfrac{2}{3}\)
\(=\dfrac{4}{3}\)
b/\(\dfrac{7}{2}+\dfrac{29}{12}-\dfrac{11}{12}\)
\(=\dfrac{7}{2}+\left(\dfrac{29}{12}-\dfrac{11}{12}\right)\)
\(=\dfrac{7}{2}+\dfrac{3}{2}\)
\(=5\)
c/\(\dfrac{26}{25}-\dfrac{3}{5}-\dfrac{2}{5}\)
\(=\dfrac{26}{25}-\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\)
\(=\dfrac{26}{25}-1\)
\(=\dfrac{26}{25}-\dfrac{25}{25}\)
\(=\dfrac{1}{25}\)
#Đang Bận Thở

Sao trong biểu thức B lại có dấu " </"?

Xem lại đề