2:

a: x=2,4-0,4=2

b: =>2x=-1,5+0,8=-0,7

=>x=-0,35

c: =>x-16=-15

=>x=1

a: =382-282+531-331

=100+200=300

b: =(7-8)+(9-10)+...+(2009-2010)

=(-1)+(-1)+....+(-1)

=-1002

c: =-(1+2+3+...+2009+2010)

=-2010*2011/2=-2021055

Bài 1: 

a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)

b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)

 đè chư a rõ

a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)

b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)

c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)

a,Gọi tổng trên là A.

Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)

b,Gọi tổng trên là B

Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)

\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)

Đặt A=B*C

B=\(\frac{24\cdot47-23}{24+47-23}=\frac{1128-23}{71-23}=\frac{1105}{48}\)

C=\(\frac{3\cdot\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\cdot\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{1}{3}\)

Suy ra A =\(\frac{1105}{144}\)

`@` `\text {Ans}`

`\downarrow`

`a.`

`A=(1/2-7/13-1/3)+(-6/13+1/2+1 1/3)`

`= 1/2 - 7/13 - 1/3 - 6/13 + 1/2 + 1 1/3`

`= (1/2 + 1/2) + (-7/13 - 6/13) + (-1/3 + 1 1/3) `

`= 1 - 1 + 1`

`= 1`

`b.`

`B=0,75+2/5+(1/9-1 1/2+5/4)`

`= 3/4 + 2/5 + 1/9 - 3/2 + 5/4`

`= (3/4+5/4)+ 1/9 + 2/5 - 3/2`

`= 2 + 1/9 - 11/10`

`= 19/9 - 11/10`

`= 91/90`

`c.`

`(-5/9).3/11+(-13/18).3/11`

`= 3/11*[(-5/9) + (-13/18)]`

`= 3/11*(-23/18)`

`= -23/66`

`d.`

`(-2/3).3/11+(-16/9).3/11`

`= 3/11* [(-2/3) + (-16/9)]`

`= 3/11*(-22/9)`

`= -2/3`

`e.`

`(-1/4).(-2/13)-7/24.(-2/13)`

`= (-2/13)*(-1/4-7/24)`

`= (-2/13)*(-13/24)`

`= 1/12`

`f.`

`(-1/27).3/7+(5/9).(-3/7)`

`= 3/7*(-1/27 - 5/9)`

`= 3/7*(-16/27)`

`= -16/63`

`g.`

`(-1/5+3/7):2/11+(-4/5+4/7):2/11`

`=[(-1/5+3/7)+(-4/5+4/7)] \div 2/11`

`= (-1/5+3/7 - 4/5 + 4/7) \div 2/11`

`= [(-1/5-4/5)+(3/7+4/7)] \div 2/11`

`= (-1+1) \div 2/11`

`= 0 \div 2/11 = 0`

`a, 4/9 + 2/5 + 5/9= (4/9 +5/9) + 2/5= 1 + 2/5= 5/5 + 2/5= 7/5`

`b, 6/13 + 5/8 + 7/13=(6/13 + 7/13) + 5/8=1+5/8= 8/8 + 5/8= 13/8`

`c, 5/11 + 6/25 + 66/121 = ( 5/11 + 66/121 ) + 6/25=1 + 6/25 = 25/25 + 6/25 = 31/25`

`d, 1/2 + 1/3 + 1/4 + 1/6 + 1/12 + 1/24= (1/2 + 1/4) +(1/3 + 1/6) + (1/12 + 1/24)= 3/4 + 1/2 + 1/8=  6/8 + 4/8 + 1/8= 11/8`

`@ yl`