giúp mình bài này với

bạn nam đọc 1 quyển sách trong 3 ngày . ngày 1 đọc 1 phần 3 số trang . ngày thứ 2 đọc 2 phần 5 số trang còn lại .ngày 3 đọc 36 trang còn lại. hỏi quyển sách có bao nhiêu trang

mình giúp bạn thì bạn giúp mình nhé

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

b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)

c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)

d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)

giúp tui

\(\dfrac{9}{14}-\dfrac{5}{8}+\dfrac{3}{7}=\dfrac{36}{56}-\dfrac{35}{56}+\dfrac{24}{56}=\dfrac{25}{56}\)

\(\dfrac{2}{9}-\dfrac{7}{10}+\dfrac{4}{5}=\dfrac{20}{90}-\dfrac{63}{90}+\dfrac{72}{90}=\dfrac{29}{90}\)

\(\dfrac{8}{19}+\dfrac{7}{5}-\dfrac{9}{20}+\dfrac{11}{12}=\dfrac{8}{19}+\dfrac{84}{60}-\dfrac{27}{60}+\dfrac{55}{60}=\dfrac{8}{19}+\dfrac{112}{60}=\dfrac{8}{19}+\dfrac{28}{15}=\dfrac{652}{285}\)

9/14-5/8+3/7=36/56-35/56+24/56=25/56

2/9-7/10+4/5=20/90-63/90+72/90=29/90

8/19+7/5-9/20+11/12

=480/1140+1596/1140-513/1140+1045/1140=2608/1140

giúp tui với

\(-\dfrac{11}{12}+\dfrac{7}{8}+\dfrac{1}{10}\)

\(=-\dfrac{1}{24}+\dfrac{1}{10}\)

\(=\dfrac{7}{120}\)

\(-\dfrac{11}{12}+\dfrac{7}{8}+\dfrac{1}{10}=-\dfrac{880}{960}+\dfrac{840}{960}+\dfrac{96}{960}=\dfrac{56}{960}=\dfrac{7}{120}\)

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

Các bạn giải giúp mình nha😐

Cần chi tiết hum bẹn

a) = 76/35 ; 37/9

b) 143/40 ; 19/15

c) 2/3 ; 42/5

d) 21/4 ; 4/7 

a, 11/7 + 3/5

= \(\dfrac{11x5}{7x5}+\dfrac{3x7}{5x7}=\dfrac{55}{35}+\dfrac{21}{35}=\dfrac{76}{35}\)

4 + 1/9

= \(\dfrac{4}{1}+\dfrac{1}{9}=\dfrac{4x9}{1x9}+\dfrac{1x1}{9x1}=\dfrac{36}{9}+\dfrac{1}{9}=\dfrac{37}{9}\)

b, 21/5 - 5/8

= \(\dfrac{21x8}{5x8}-\dfrac{5x5}{8x5}=\dfrac{168}{40}-\dfrac{25}{40}=\dfrac{143}{40}\)

2 - 11/15

=\(\dfrac{2}{1}-\dfrac{11}{15}=\dfrac{30}{15}-\dfrac{11}{15}=\dfrac{19}{15}\)

MIK LÀM ĐC NGẦN ĐẤY THUI BẸN THÔNG CẢM

a, \(\dfrac{38}{7}\) + ( \(\dfrac{16}{7}\) - \(\dfrac{5}{3}\))

= \(\dfrac{38}{7}\) + \(\dfrac{16}{7}\) - \(\dfrac{5}{3}\)

= \(\dfrac{54}{7}\) - \(\dfrac{5}{3}\)

= \(\dfrac{162}{21}\) - \(\dfrac{35}{21}\)

= \(\dfrac{127}{21}\)

b, \(\dfrac{29}{9}\) - ( \(\dfrac{14}{9}\) + \(\dfrac{11}{9}\) )

= \(\dfrac{29-14-11}{9}\)

= \(\dfrac{4}{9}\)

c, ( \(\dfrac{2}{7}\) + \(\dfrac{3}{8}\) + \(\dfrac{12}{7}\) + \(\dfrac{5}{8}\) ) - \(\dfrac{19}{63}\)

= ( \(\dfrac{2+12}{7}\) + \(\dfrac{3+5}{8}\)) - \(\dfrac{19}{63}\)

= \(\left(1+2\right)\) - \(\dfrac{19}{63}\)

= \(3-\dfrac{19}{63}\)

= \(\dfrac{189}{63}\) - \(\dfrac{19}{63}\)

= \(\dfrac{170}{63}\)

d, \(\dfrac{3}{7}\) . \(\dfrac{4}{9}\) + \(\dfrac{4}{9}\) . \(\dfrac{4}{7}\) - \(\dfrac{4}{9}\)

= \(\dfrac{4}{9}\) . ( \(\dfrac{3}{7}\) + \(\dfrac{3}{7}\) ) - \(\dfrac{4}{9}\)

= \(\dfrac{4}{9}\) . \(1-\dfrac{4}{9}\)

= \(\dfrac{4}{9}\) - \(\dfrac{4}{9}\)

= 0

chú ý dấu chấm là dấu nhân

a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)

\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)

\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)

=0

b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)

\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)

\(=\dfrac{7}{3}\cdot0=0\)

c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)

\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)

d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)

\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)

\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)

\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)

\(=0:\dfrac{3}{8}=0\)

\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)

\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)

\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)

\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)

\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)

\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)

\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)

a) Dấu hiệu là điểm bài thi học kì của 100 học sinh lớp 7 của một trường Trung học Cơ Sở Hòa Bình. Số các dấu hiệu là 100
b) Bảng tần số
 

Nhận xét: Giá trị lớn nhất là 19, giá trị nhỏ nhất là 1; tần số lớn nhất là 13, tần số nhỏ nhất là 1.