Bài 1

1) (-35) + (-9)

= -(35 + 9)

= -44

2) (-17) + (-53)

= -(17 + 53)

= -70

3) (-7) + (-14)

= -(7 + 14)

= -21

4) (-26) + (-6)

= -(26 + 6)

= -32

5) (-75) + (-50)

= -(75 + 50)

= -125

6) (-5) + (-248)

= -(5 + 248)

= -253

7) (-37) + (-23)

= -(37 + 23)

= -60

8) (-25) + (-15)

= -(25 + 15)

= -40

9) (-36) + (-6)

= -(36 + 6)

= -42

Bài 2

1) (-65) + 37

= -(65 - 37)

= -28

2) (-136) + 122

= -(136 - 122)

= -14

3) 19 + (-49)

= -(49 - 19)

= -30

4) 21 + (-78)

= -(78 - 21)

= -57

5) 363 + (-400)

= -(400 - 363)

= -37

6) (-15) + 7

= -(15 - 7)

= -8

7) (-237) + 137

= -(237 - 137)

= -100

8) 78 + (-123)

= -(123 - 78)

= -45

9) (-75) + 50

= -(75 - 50)

= -25

Bài 3

1) (-73) + 99

= 99 - 73

= 26

2) (-45) + 83

= 83 - 45

= 38

3) (-125) + 147

= 147 - 125

= 22

4) 121 + (-21)

= 121 - 21

= 100

5) 1945 + (-945)

= 1945 - 945

= 1000

6) 19 + (-13)

= 19 - 13

= 6

7) (-15) + 27

= 27 - 15

= 12

8) 65 + (-17)

= 65 - 17

= 48

9) (-73) + 122

= 122 - 73

= 49

a) A = \(\sum\limits^{50}_1\left(2x\right)-\sum\limits^{50}_1\left(2x-1\right)\) = 5050

b) B = \(\sum\limits^{2010}_1x^3\) = 4084663313000

Giải :

\(\dfrac{x+11}{89}+\dfrac{x+13}{87}-\dfrac{x+15}{85}-\dfrac{x+17}{83}=0\\ =>\left(\dfrac{x+11}{89}+1\right)+\left(\dfrac{x+13}{87}+1\right)-\left(\dfrac{x+15}{85}+1\right)-\left(\dfrac{x+17}{83}+1\right)=0\\ =>\left(\dfrac{x+11+89}{89}\right)+\dfrac{x+13+87}{87}-\dfrac{x+15+85}{85}-\dfrac{x+17+83}{83}=0\\ =>\dfrac{x+100}{89}+\dfrac{x+100}{87}-\dfrac{x+100}{85}-\dfrac{x+100}{83}=0\\ =>\left(x+100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\right)=0\\ =>\left[{}\begin{matrix}x+100=0\\\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}=0\left(voli\right)\end{matrix}\right.=>x=-100\)

  tìm x, y, z biết 

\(2x=3y\\ =>\dfrac{x}{3}=\dfrac{y}{2}\\ 4y=5z\\ =>\dfrac{y}{5}=\dfrac{z}{4}\\ \dfrac{x}{3}=\dfrac{y}{2}\\ =>\dfrac{x}{3.5}=\dfrac{y}{2.5}\\ =>\dfrac{x}{15}=\dfrac{y}{10}\\ \dfrac{y}{5}=\dfrac{z}{4}\\ =>\dfrac{y}{5.2}=\dfrac{z}{4.2}\\ =>\dfrac{y}{10}=\dfrac{z}{8}\\ =>\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\\ =>\left\{{}\begin{matrix}x=\dfrac{1}{3}.15=5\\y=\dfrac{1}{3}.10=\dfrac{10}{3}\\z=\dfrac{1}{3}.8=\dfrac{8}{3}\end{matrix}\right.\)

  tìm x, y, z biết

Giải

\(\dfrac{x}{3}=\dfrac{y}{4}\\ \Leftrightarrow\dfrac{x}{3.3}=\dfrac{y}{4.3}\\\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\\ \dfrac{y}{3}=\dfrac{z}{5}\\ \Leftrightarrow \dfrac{y}{3.4}=\dfrac{z}{5.4}\\ \Leftrightarrow\dfrac{y}{12}=\dfrac{z}{20}\\ =>\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau 

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x+3x+z}{2.9+3.12+20}=\dfrac{6}{74}=\dfrac{3}{37}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{37}\times9=\dfrac{27}{37}\\y=\dfrac{3}{37}\times12=\dfrac{36}{37}\\z=\dfrac{3}{37}\times20=\dfrac{60}{37}\end{matrix}\right.\)