Bài 2:

n) Ta có: \(N=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2014\cdot2016}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2014\cdot2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)

\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)\)

\(=2\cdot\dfrac{1007}{2016}=\dfrac{1007}{1008}\)

o) Ta có: \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+\dfrac{1}{9\cdot12}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

a) Ta có: \(\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(=\dfrac{58}{9}+\dfrac{40}{11}-\dfrac{40}{9}\)

\(=2+\dfrac{40}{11}=\dfrac{62}{11}\)

Bài 2:

b) Ta có: \(10\dfrac{1}{5}-5\dfrac{1}{2}\cdot\dfrac{60}{11}+3:15\%\)

\(=\dfrac{51}{5}-\dfrac{11}{2}\cdot\dfrac{60}{11}+3:\dfrac{3}{20}\)

\(=\dfrac{51}{5}-30+20\)

\(=\dfrac{51}{5}-10=\dfrac{1}{5}\)

c) Ta có: \(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{97\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

Bài 2: 

d) Ta có: \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}\cdot\dfrac{-7}{11}\)

\(=-\dfrac{5}{11}\)

e) Ta có: \(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\dfrac{12}{7}-\dfrac{5}{7}=1\)

a) Ta có: \(19\dfrac{5}{8}:\dfrac{7}{12}+\left(-15\dfrac{1}{4}\right):\dfrac{7}{12}\)

\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)

\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)

                             Bây giờ ai đã quên chưa ?

                    Màu hoa phượng nở khi hè vừa sang

                             Bâng khuâng dưới ánh trăng vàng

                    Tặng nhau cánh phượng ai mang đi rồi 

                              Ngày xưa chỉ có vậy thôi 

                     Có ai biết đc để rồi cách xa

                              Mùa hè từng mùa qua

                      Tiếc hoài cái tuổi ngọc ngà chẳng quên

                              Nỗi buồn ko thể đặt tên

                      Nhẹ nhàng nhưng lại mênh mông trong lòng

                              Ai còn nhớ kỉ niệm ko ?

                      Ngày xưa , một cánh phượng hồng đã chao .

k cho mik nha!

các bạn có thể yuwj nghĩ ra giúp mình đc ko

2.2.2.3.3=2³.3²=8.9=72

100.10.10.10=10.10.10.10.10=10⁵=100000

2 . 2 . 2 . 3 . 3 = 2³ . 3²

100 . 10 . 10 . 10 = 10 . 10 . 10 . 10 . 10 = 10⁵

Thanks mọi người nhiều!

\(B=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)

\(B=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}\)

\(B=-\left(\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\right)\)

\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)

\(B=-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)

\(B=-\left(\frac{1}{10}.\frac{11}{2}\right)\)

\(B=\frac{-11}{20}< \frac{-11}{21}\)

Vậy \(B< \frac{-11}{21}\)

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To balance my productive and leisure activities, I made sure to allocate some time for relaxation as well. I enjoyed watching movies, spending time with friends, and exploring new places in my city. These moments gave me a chance to unwind and create beautiful memories with my loved ones.

Overall, my summer break was a fulfilling one. It allowed me to bond with my family, improve my English proficiency, explore my interests, and take moments to relax. I am thankful for this wonderful break, as it has prepared me for the challenges that lie ahead in 10th grade.

70:

a: Quãng đường đi được là:

1,5(25,5+2,5)=28*1,5=42km

b: Quãng đường đi được là:

1,5*(25,5-2,5)=23*1,5=31,5km

69

Đổi 4 giờ 12 phút = 4,2 giờ

Vận tốc khi nước yên tĩnh là:

\(\dfrac{63}{4,2}=15\) (km/giờ)

Vận tốc ca nô khi xuôi dòng là:

\(15+6=21\) (km/giờ)

Vận tốc ca nô khi ngược dòng là:

\(15-6=9\) (km/giờ)

68

a)Đổi 50 phút = 5/6 h

Xe máy đi trước ô tô: 32,4 * 5/6 = 27 (km)

Thời gian cả 2 xe cùng đi là:

(162-27):(32,4+48,6) = 135 : 81 = 5/3 (giờ) = 1h40'

Hai xe gặp nhau lúc: 7h30' + 50' + 1h40' = 10h

b) Điểm gặp nhau cách A: 32,4(5/6 + 5/3) = 32,4 * 5/2 = 81(km)

Đáp số: a) 10h

b)81km

Có: ⎧⎩⎨⎪⎪⎪⎪⎪⎪|x−1|≥x−1|x−2|≥x−2|x−3|≥3−x|x−4|≥4−x{|x−1|≥x−1|x−2|≥x−2|x−3|≥3−x|x−4|≥4−x với mọi x

Do đó, D≥(x−1)+(x−2)+(3−x)+(4−x)D≥(x−1)+(x−2)+(3−x)+(4−x)

hay D≥4D≥4

Dấu "=" xảy ra khi {x−2≥03−x≤0{x−2≥03−x≤0⇒{x≥2x≤3⇒{x≥2x≤3⇒2≤x≤3⇒2≤x≤3

Vậy GTNN của C là 4 khi 

Huy tú :mình ko hiểu bạn làm gì cả. hình như bạn copy thì phải .

có         với mọi x lla sao bạn. bạn tư chế à

Có: ⎧⎩⎨⎪⎪⎪⎪⎪⎪|x−1|≥x−1|x−2|≥x−2|x−3|≥3−x|x−4|≥4−x{|x−1|≥x−1|x−2|≥x−2|x−3|≥3−x|x−4|≥4−x với mọi x

Do đó, D≥(x−1)+(x−2)+(3−x)+(4−x)D≥(x−1)+(x−2)+(3−x)+(4−x)

hay D≥4D≥4

Dấu "=" xảy ra khi {x−2≥03−x≤0{x−2≥03−x≤0⇒{x≥2x≤3⇒{x≥2x≤3⇒2≤x≤3⇒2≤x≤3

Vậy GTNN của C là 4 khi 2≤x≤3

xl mik ko cs sách đó