\(P=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(P=\frac{1}{1}-\frac{1}{100}\)

\(P=\frac{99}{100}\)

\(HT\)

\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)

\(P=1+0+0+....+0+\dfrac{-1}{100}\)

\(P=1+\dfrac{-1}{100}\)

\(P=\dfrac{99}{100}\)

\(\frac{6}{5}< \frac{x}{10}< \frac{3}{2}\)

\(\Leftrightarrow\frac{12}{10}< \frac{x}{10}< \frac{15}{10}\)

\(\Leftrightarrow12< x< 15\)

\(\Leftrightarrow x\in\left\{13;14\right\}\)

Vậy....

13 và 14 nhé!

N=99957

N=99968

ok

bye

Bài 3

Số khẩu trang lớp 6A ủng hộ là : \(500\times\frac{2}{5}=200\text{ hộp}\)

Số khẩu trang lớp 6B ủng hộ là : \(120:\frac{3}{4}=160\text{ hộp}\)

Số khẩu trang lớp 6C ủng hộ là : \(500-200-160=140\text{ hộp}\)

ủa toán mà

Số học sinh giỏi : 50 x 7/50 = 7 học sinh

số học sinh kém : 50 x 1/50 = 1 học sinh 

=\(\dfrac{28.3}{15.16}+\left(\dfrac{8}{15}-\dfrac{69.5}{60.23}\right).\dfrac{-54}{51}\)

=\(\dfrac{7}{20}+\left(\dfrac{8}{15}-\dfrac{1}{6}\right).\dfrac{-54}{51}\)

=\(\dfrac{7}{20}+\left(\dfrac{16}{30}-\dfrac{5}{30}\right).\dfrac{-54}{51}\)

=\(\dfrac{7}{20}+\dfrac{11}{30}.\dfrac{-54}{51}\)

=\(\dfrac{7}{20}+\dfrac{-99}{255}\)

=\(\dfrac{357}{1020}+\dfrac{-396}{1020}\)

=\(\dfrac{-39}{1020}\)