xin lỗi đây là bài tính hợp lí nhé 

-3/11.(-22)/66.121/15

=(-3).(-22).121

 11.66.15

=11

15

3/7.2/5.7/3.20.19/72

=3.2.7.20.19

7.5.3.72

=76

16

6/7.8/13+6/13.9/7-3/13.6/7

=6/7.8/13+6/7.9/13-3/13.6/7

=6/7.(8/13+9/13-3/13)

=6/7.14/13

=12/13

-1/4.152/11+68/4.(-1)/11

=152/4.(-1)/11+68/4.(-1)/11

=(-1)/11.(152/4+68/4)

=(-1)/11.220/4

=-110/22

-5/7.2/11+(-5)/7.9/11+12/7

=-5/7.2/11+-5/7.9/11+12/7

=-5/7.(2/11+9/11)+12/7

=-5/7.1+12/7

=(-5)/7+12/7

=7/7

=1

146/13-(18/7+68/13)

=146/13-18/7-68/13

=(146/13-68/13)-18/7

=78/13-18/7

=6-18/7

=42/7-18/7

=24/7

\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)

\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)

\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)

A =\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)

   = \(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.11}+\frac{1}{9.11}-\frac{1}{11.13}\)

   = \(\frac{1}{1.3}-\frac{1}{11.13}\)

   = \(\frac{1}{3}-\frac{1}{143}\)

  = \(\frac{140}{429}\)

Dấu "." là dấu nhân

\(\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}\)

\(=\dfrac{67}{9}+\dfrac{51}{11}-\dfrac{31}{9}\)

\(=\dfrac{67}{9}-\dfrac{31}{9}+\dfrac{51}{11}\)

\(=4+\dfrac{51}{11}\)

\(=\dfrac{95}{11}\)

Chúc bạn học tốt

jjjjjjjjjj

\(C=\frac{4}{9.11.13}+\frac{4}{11.13.15}+...+\frac{4}{59.61.63}\)

\(=2\left(\frac{2}{9.11.13}+\frac{2}{11.13.15}+...+\frac{2}{59.61.63}\right)\)

\(=2\left(\frac{1}{9.11}-\frac{1}{11.13}+\frac{1}{11.13}-\frac{1}{13.15}+...+\frac{1}{59.61}-\frac{1}{61.63}\right)\)

\(=2\left(\frac{1}{9.11}-\frac{1}{61.63}\right)\)

\(=2.\frac{416}{42273}\)

\(=\frac{832}{42273}\)

\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^{19}}\)

\(\Leftrightarrow\)\(\frac{3^2.\left(2^2\right)^2.2^{16.2}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^{19}}\)

\(\Leftrightarrow\)\(\frac{2^4.2^{32}.3^2}{11.2^{35}-2^{76}}\)

\(\Leftrightarrow\)\(\frac{2^{36}.3^2}{2^{35}.\left(11-2^{41}\right)}\)

\(\Leftrightarrow\)\(\frac{2.3^2}{11-2^{41}}\)

Hết biết giải rồi 

\(\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}=\frac{3^2.2}{9}=\frac{3^2.2}{3^2}=2\)

\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\)\(\frac{3^2.2^{2.2}.2^{16.2}}{11.2^{13}.2^{11.2}-2^{4.9}}\)

\(=\)\(\frac{2^{36}.3^2}{11.2^{35}-2^{36}}\)

\(=\)\(\frac{2^{36}.3^2}{2^{35}.\left(11-2\right)}\)

\(=\)\(\frac{2.3^2}{9}\)

\(=\)\(\frac{2.3^2}{3^2}\)

\(=\)\(2\)

có cần cóp nguyên bài thế k

=1 nha ban

Ta có : 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12+......+ 2006 - 2007 - 2008 + 2009 

= (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) + (9 + 10 - 11 - 12)+...... + (2005 + 2006 - 2007 - 2008) + 2009 

= -4 + -4 + ...... + (-4) + 2009

= (-4) x 502 + 2009

= (-2008) + 2009

= 1