2

2

2

1 gio 8 phut

10 giờ 12 phút : 9 = 1 giờ 8 phút

Ý nghĩa: Nâng cao tinh thần đấu tranh chống quân xâm lược của dân tộc.

Gọi \(x\) là số học sinh giỏi lớp 5A. Theo đề bài ta có:

\(x\) x \(\dfrac{3}{5}\) = 6 ⇒ \(x\) = 6 x \(\dfrac{5}{3}\) = 10 (học sinh)

Vậy lớp 5A có 10 học sinh giỏi

Số học sinh lớp 5A là:

            6 : \(\dfrac{3}{5}\)\frac{3}{5}

​= 10 ( em)

                     Đ/S: 10 em

Số học sinh giỏi của lớp 5A:

6:3/5=10(HS)

ĐS: 10HS

Bn cho đáp án cho sắn đi, mk trả lời cho lẹ!!

dễ ợt thứ 5

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

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

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

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

\(\Rightarrow5^2\cdot5^{x+3}=5^6\\ \Rightarrow5^{x+3}=5^4\\ \Rightarrow x+3=4\\ \Rightarrow x=1\)

\(25.5^{x+3}=5^6\)

\(\Rightarrow5^2.5^{x+3}=5^6\)

\(\Rightarrow5^{x+3}=5^6:5^2\)

\(\Rightarrow5^{x+3}=5^4\)

\(\Rightarrow x+3=4\)

\(\Rightarrow x\)       \(=4-3\)

\(\Rightarrow x\)       \(=1\)

\(=\left(x-x\right)^2+\left(y-y\right)^2=0^2+0^2=1+1=2\)

\(x^2-x-y^2-y\)

\(=\left(x^2-y^2\right)-\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

đề bài là gì thế bn???