a,-5/9 + 8/18 + -2/11 + 4/-9 + 7/15                                            b,5/17 + -9/15 + -2/17 + 2/5 + 14/17

= -5/9 + 4/9 + -2/+ 4/-9 + 7/15                                                 = 5/17 + -3/5 + -2/17 + 2/5 + 14/17

=(-5/9 + 4/9 + -4/9 ) +-2/11 + 7/15                                            = ( 5/17 + -2/17 + 14/17) + ( 2/5 + -3/5)

= -1/3 + -2/11 + 7/15                                                               =1 + -1/5

=-75/225 + -30/ 225 + 77/225                                                  =4/5              

= -25/225 = -1/9                                                                        ( 4/5 = 0.8 )

\(A=\frac{2}{7}+\frac{-3}{8}+\frac{11}{7}+\frac{1}{3}+\frac{1}{7}+\frac{5}{-8}\)

\(A=\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)+\left(\frac{-3}{8}+\frac{5}{-8}\right)+\frac{1}{3}\)

\(A=2-1+\frac{1}{3}\)

\(A=\frac{4}{3}\)

\(B=\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+17\)

\(B=\left(\frac{3}{17}+\frac{14}{17}\right)+\frac{-5}{13}+\frac{-18}{35}+17\)

\(B=1+\frac{-5}{13}+\frac{-18}{35}+17\)

\(B=18+\frac{-5}{13}+\frac{-18}{35}\)

\(B=\frac{7781}{455}\)

Ta có : \(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)

\(=\frac{2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{14}-\frac{1}{17}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{17}\right)=\frac{2}{3}.\frac{15}{34}=\frac{5}{17}< 1\)

nó = 2/3(3/2x5+3/5x8+...+3/14x17)

=2/3(1/2-1/5+1/5-1/8+...+1/14-1/17)

=2/3x(1/2-1/17)

=2/3x15/34=5/17<1( điều phải chứng minh)

a)1-2-3+4+5-6-7+8+.....+97-98-99+100

=(1-2)+(-3+4)+5-6+...+(97-98)+(-99+100)

=-1+(-1)+(-1)+...+(-1)+(-1) có 50 cap nen =(-1).50 =-50

1) \(\left(+15\right)+\left(+17\right)=15+17=32\)

2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)

3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)

4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)

5) \(\left(-15\right)+20=20-15=5\)

6) \(\left(-5\right)+8+7+5\)

\(=\left(-5+5\right)+\left(8+7\right)\)

\(=15\)

7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)

\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)

\(=\left(-10\right)+\left(-11\right)\)

\(=-21\)

8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)

\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)

\(=10+\left(-30\right)\)

\(=-20\)

9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)

\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)

\(=\left(-17\right)+\left(-100\right)\)

\(=-117\)

10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)

\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)

\(=\left(-400\right)+40+240\)

\(=\left(-360\right)+240\)

\(=-120\)

11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)

\(=\left(-596\right)+2001+1999-404+189\)

\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)

\(=\left(-1000\right)+2190+1999\)

\(=1190+1999\)

\(=3189\)

12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)

\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)

\(=\left(378+121\right)+\left(-400\right)+218\)

\(=499-400+218\)

\(=99+218\)

\(=317\)

\(\text{#}Toru\)

a) 12 + 3 = 15     15 + 4 = 19     8 + 2 = 10     14 + 3 = 17

15 - 3 = 12     19 - 4 = 15     10 - 2 = 8     17 - 3 = 14

b) 11 + 4 + 2 = 17     19 - 5 - 4 = 10     14 + 2 - 5 = 11

a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)

Vậy A<\(\frac{1}{2}\).

b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                  \(\frac{1}{3^2}< \frac{1}{2.3}\)

                  \(\frac{1}{4^2}< \frac{1}{3.4}\)

                    ...

                   \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

Vậy \(B< 1\).

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