\(A=\frac{-3}{5}+\left(\frac{-2}{5}+2\right)\)

\(A=\frac{-3}{5}+\frac{8}{5}\)

\(A=1\)

\(B=\frac{3}{7}+\left(\frac{-1}{5}+\frac{-3}{7}\right)\)

\(B=\frac{3}{7}+\frac{-1}{5}+\frac{-3}{7}\)

\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\frac{-1}{5}\)

\(B=0+\frac{-1}{5}\)

\(B=\frac{-1}{5}\)

\(C=\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+\frac{52}{9}\)

\(C=\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)

\(C=\frac{-7}{9}.1+\frac{52}{9}\)

\(C=\frac{-7}{9}+\frac{52}{9}\)

\(C=5\)

\(D=50\%.\frac{4}{3}.10.\frac{7}{35}.0,75\)

\(D=\frac{50}{100}.\frac{4}{3}.10.\frac{7}{35}.\frac{75}{100}\)

\(D=\frac{1}{2}.\frac{4}{3}.10.\frac{1}{5}.\frac{3}{4}\)

\(D=\left(\frac{1}{2}.\frac{1}{5}.10\right).\left(\frac{4}{3}.\frac{3}{4}\right)\)

\(D=1.1\)

\(D=1\)

A= 1

B= -1/5

C=5

D=1

mik có ý kiến lần sau bạn đặt câu hỏi nhớ ghi phân số trên dưới cho dễ nhìn nha

Các bạn làm chi tiết đc ko!!

\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)

\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)

\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)

\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)

\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)

a,  -3/4 + 3/7 + -1/4 + 4/9 + 4/7

=

=

=

b. \(\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+\dfrac{52}{9}\)

= \(\dfrac{-7}{9}.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}\)

= \(\dfrac{-7}{9}.1+\dfrac{52}{9}\)

= \(\dfrac{-7}{9}+\dfrac{52}{9}\)

= \(\dfrac{45}{9}\)

Chúc bạn học tốt!! ^^

\(a\)) \(\dfrac{-3}{4}+\dfrac{3}{7}+\dfrac{-1}{4}+\dfrac{4}{9}+\dfrac{4}{7}\)

\(=\left(\dfrac{-3}{4}+\dfrac{-1}{4}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{4}{9}\)

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

\(=0+\dfrac{4}{9}=\dfrac{4}{9}\)

\(a,\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)

\(=\frac{-3}{4}+\frac{-1}{4}+\frac{3}{7}+\frac{4}{7}+\frac{4}{9}\)

\(=-1+1+\frac{4}{9}\)

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

\(b,\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)

\(=\frac{-7}{9}\cdot1+\frac{52}{9}\)

\(=\frac{-7}{9}+\frac{52}{9}=\frac{45}{9}=5\)

Ý c) bn tự làm nha. Mk ko làm đc. Nhìn nó rắc rối lắm.

Chúc bn hc tốt nha !!! TỨ DIỆP THẢO

câu c bn phá ngoặc ra rồi ghép lại với nhau là xong

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

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

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

=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)

=\(\dfrac{13}{10}\)

\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)

=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\) 

=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\) 

=\(-\dfrac{7}{25}-\dfrac{18}{25}\) 

=\(-\dfrac{25}{25}\) = \(-1\)

(-3,2) +\(\dfrac{15}{64}+\left(0.8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\) 

=\(-\dfrac{16}{5}+\dfrac{15}{64}+(\dfrac{4}{5}-\dfrac{34}{15}):\dfrac{11}{3}\)

=\(-\dfrac{16}{5}+\dfrac{15}{64}+(-\dfrac{22}{5}):\dfrac{11}{3}\)

=\(-\dfrac{16}{5}+\dfrac{15}{64}+\dfrac{-2}{5}\)

=(\(-\dfrac{16}{5}+\dfrac{-2}{5}\)) +\(\dfrac{15}{64}\)

=\(-\dfrac{18}{5}+\dfrac{15}{64}\)

\(-\dfrac{1152}{320}+\dfrac{75}{320}\)

\(-\dfrac{1077}{320}\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)

b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)

\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)

c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)

d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)

e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)