TL

\(\frac{3}{4}+\frac{2}{3}=\frac{17}{12}\)

\(\frac{4}{5}+\frac{1}{2}-\frac{7}{10}=\frac{3}{5}\)

\(\frac{7}{6}\).\(\frac{1}{3}+\frac{7}{6}\).\(\frac{1}{2}\)-\(\frac{7}{6}\) . \(\frac{5}{8}\)=\(\frac{7}{6}.\)(\(\frac{1}{3}+\frac{1}{2}-\frac{5}{8}\))=\(\frac{7}{6}.\)\(\frac{5}{24}\)=\(\frac{35}{144}\)

HT

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

Câu a:

7/10.4/9 + 3/10.4/9 - 1 7/9

= 4/9.(7/10 + 3/10) - 16/9

= 4/9 - 16/9

= - 12/9

= - 4/3

Câu b:

- 5/6 + 4/9.(5/4 - 2/3).(-3)^2 + 5/9.30%

= - 5/6 + [5/9 - 8/27].9 + 5/9.3/10

= - 5/6 + 5 - 8/3 + 1/6

= (-5/6 + 1/6) - 8/3 + 5

= - 4/6 - 8/3 + 15/3

= - 2/3 - 8/3 + 15/3

= - 10/3 + 15/3

= 5/3

a, 13/6+5/8 : -3/4 - 7/12.4

= 13/6 + -5/6-7/3

=8/6-7/3

= -6/6

= -1

b, ( 73/5 - 21/3) + ( 4/3-43/5 )

= 73/5-21/3+4/3-43/5

=( 73/5-43/5)-(21/3-4/3)

= 6-17/3

=1/3

c, 7/5.4/9 +7/5: 9/16- 14/10.2/9

= 7/5.4/9 +7/5.16/9 - 14/45

=7/5.(4/9+16/9)-14/45

=7/5.20/9-14/45

= 140/45 - 14/45

= 126/45

Xong rùi nè! Nhưng bạn kiểm tra lại giùm nhé vì làm vào ban đêm nên hơi bất tiện

Trời ạ. 7 3/5 chứ k phải 73/5 -_-

a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)

\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)

\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)

\(=\frac{-60}{45}=\frac{-4}{3}\)

b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)

\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)

\(=\frac{-5}{6}+\frac{7}{3}+\frac{1}{6}\)

\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)

=\(=\frac{10}{6}=\frac{5}{3}\)

Câu c:

1/2017.(1/3 - 7/12 + 1/4) - (6 2/3 - 1,5.2^2) :5/6

= 1/2017.(4/12 - 7/12 + 3/12) - (20/3 - 1,5x 4) : 5/6

= 1/2017.0 - (20/3 - 6) x 6/5

= 0 - (20/3 - 18/3) x 6/5

= - 2/3 x 6/5

= - 4/5

a)(-5/7+5/7)+3/-5=3/-5

b) câu này mình chả hiểu biểu thức của bạn

c)-2/5-3/7+10/7+7/5

=5/5+7/7=2

2

a)l2x-6l=0

 th1 2x+6=0

x=3

th2 -2x-6=0

x=-3

b)-3/4-x=1/5

-x=1/5+3/4=19/20

x=-19/20

k cho mình nhé

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