Ngày thứ nhất đọc được :

   90 x 4 / 9 = 40 (trang)

Ngày thứ hai đọc được :

  90 x 2/5 = 36 (trang)

Ngày thứ ba đọc được :

  90 - 36 - 40 = 14 (trang)

\(-\dfrac{9}{16}.\dfrac{13}{3}-\left(-\dfrac{3}{4}\right)^2.\dfrac{19}{3}\\ =-\dfrac{9}{16}.\dfrac{13}{3}-\dfrac{9}{16}.\dfrac{19}{3}\\ =-\dfrac{9}{16}.\left(\dfrac{13}{3}+\dfrac{19}{3}\right)\\ =-\dfrac{9}{16}.\dfrac{32}{3}\\ =-\dfrac{3.3.16.2}{16.3}=-3.2\\ =-6\)

Điểm B ở đâu ra thế em?

3x+7 chia hết cho x

Nhận thấy : Với mọi số nguyên x thì 3x luôn chia hết cho x

Do đó để 3x+7 chia hết cho x

Thì : 7 phải chia hết cho x

=> \(x\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

cảm ơn nha

\(A=3+3^2+3^3+...+3^{199}+3^{200}\\ \Rightarrow3A=3^2+3^3+3^4+....+3^{200}+3^{201}\\ \Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{200}+3^{201}\right)-\left(3+3^2+3^3+...+3^{199}+3^{200}\right)\\ \Rightarrow2A=3^{201}-3\\ \Rightarrow2A+3=3^{201}\)( Là một lũy thừa của 3 ) => DPCM

cảm ơn

\(a.\dfrac{31}{17}+\left(-\dfrac{5}{13}\right)+\left(-\dfrac{8}{13}\right)-\dfrac{4}{17}\\ =\left(\dfrac{31}{17}-\dfrac{4}{17}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\\ =\dfrac{27}{17}+\left(-1\right)\\ =\dfrac{27}{17}+\dfrac{-17}{17}=\dfrac{10}{17}\\ b.\left(-2\right)^3-1=-8-1=-9\\ \dfrac{5}{27}\cdot\left(\dfrac{-3}{2}\right)^3=\dfrac{5}{27}\cdot\dfrac{-27}{8}=-\dfrac{5}{8}\)

a: \(\dfrac{3}{5}+\dfrac{3}{5\cdot9}+...+\dfrac{3}{97\cdot101}\)

\(=\dfrac{3}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{97\cdot101}\right)\)

\(=\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{4}\cdot\dfrac{100}{101}=\dfrac{75}{101}\)

b: \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

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

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

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

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

c: \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)

d: \(\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\cdot...\cdot\left(1-\dfrac{2014}{7}\right)\)

\(=\left(1-\dfrac{7}{7}\right)\cdot\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot...\cdot\dfrac{-2007}{7}\)

\(=\left(1-1\right)\cdot\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot...\cdot\dfrac{-2007}{7}\)

=0

\(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

=>\(x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(x+\dfrac{8}{45}=\dfrac{29}{45}\)

=>\(x=\dfrac{29-8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)

Bài 6:

Số đường thẳng là: \(4\cdot\dfrac{3}{2}=2\cdot3=6\left(đường\right)\)

Cứ 1 điểm sẽ tạo với 4 - 1 điểm còn lại  4 - 1 tia

Với 4 điểm ta sẽ tạo được số tia là:

(4 -  1) x 3  = 12 (tia)

Kết luận có 12 tia có gốc là một trong 4 điểm đã cho đó lần lượt là các tia:

EF; EG; EH; FE; FG; FH; GE; GF; GH; HE; HF; HG

a: \(1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}\)

\(=\dfrac{30-10-5-3-2}{30}\)

\(=\dfrac{10}{30}=\dfrac{1}{3}\)

b: \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=0\)

c: \(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{27}-\left(+\dfrac{7}{18}\right)+\dfrac{4}{35}-\left(-\dfrac{2}{7}\right)\)

\(=\dfrac{-1}{2}+\dfrac{3}{5}+\dfrac{-1}{9}+\dfrac{1}{27}-\dfrac{7}{18}+\dfrac{4}{35}+\dfrac{2}{7}\)

\(=\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{1}{27}\)

\(=\dfrac{21+4+10}{35}+\dfrac{-9-2-7}{18}+\dfrac{1}{27}\)

\(=\dfrac{35}{35}-\dfrac{18}{18}+\dfrac{1}{27}=\dfrac{1}{27}\)

d: \(\dfrac{3}{5}+\dfrac{3}{11}-\left(-\dfrac{3}{7}\right)+\left(\dfrac{2}{17}\right)-\dfrac{1}{35}-\dfrac{3}{4}+\left(-\dfrac{23}{44}\right)\)

\(=\left(\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{1}{35}\right)+\left(\dfrac{3}{11}-\dfrac{3}{4}-\dfrac{23}{44}\right)+\dfrac{2}{17}\)

\(=\dfrac{21+15-1}{35}+\dfrac{12-33-23}{44}+\dfrac{2}{17}\)

\(=\dfrac{35}{35}-\dfrac{44}{44}+\dfrac{2}{17}=\dfrac{2}{17}\)

cam on nha