1) 5 + (-4) = 1

2) (-8) + 2 = -6

3) 8 + (-2) = 6

4) 11 + (-3) = 8

5) (-11) + 2 = -9

6) (-7) + 3 = -4

7) (-5) + 5 = 0

8) 11 + (-12) = -1

9) (-18) + 20 = 2

10) (15) + (-12) = 3

11) (-17) + 17 = 0

12) 16 + (-2) = 14

13) (30) + (-14) = 16

14) (-19) + 20 = 1

15) (-18) + 15 = -3

16) (10) + (-6) = 4

17) (-28) + 14 = -14

18) 15 + (-30) = -15

19) (15) + (-4) = 11

20) (-21) + 11 = -10

21) 8 + (-22) = -14

22) (-15) + 4 = -11

23) (-3) + 2 = -1

24) 17 + (-14) = 3

25) 17 + (-14) = 3

thế này đọc đc mới đỉnh

bai EZ​ qua

????

:))))

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

1, 3/5 + -4/15 = \(\dfrac{9}{15}\)+ \(\dfrac{-4}{15}\)= \(\dfrac{5}{15}\)= \(\dfrac{1}{3}\)

2, -1/3 + 2/5 + 2/15= \(\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}\)=\(\dfrac{3}{15}=\dfrac{1}{5}\)

3, -3/5 + 7/21 + -4/5 + 7/5= \(\dfrac{1}{3}\)

\(1.\dfrac{3}{5}+\dfrac{-4}{15}=\dfrac{9}{15}+\dfrac{-4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

\(2.\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{2}{15}=\dfrac{-5}{15}+\dfrac{6}{15}+\dfrac{2}{15}=\dfrac{3}{15}=\dfrac{1}{5}\)

\(3.\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}=\left(\dfrac{-3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}=0+\dfrac{7}{21}=\dfrac{1}{3}\)

^{Đã Làm !?}

\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)

a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)

b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)

c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)

=98

[1-(3/4-2/3)] - [1-(5/3-1/4)] - [1-(4/3+3/4)]

\(=\left[1-\left(\frac{3}{4}-\frac{2}{3}\right)\right]-\left[1-\left(\frac{5}{3}-\frac{1}{4}\right)\right]-\left[1-\left(\frac{4}{3}+\frac{3}{4}\right)\right]\)

\(=\left[1-\frac{1}{12}\right]-\left[1-\frac{17}{12}\right]-\left[1-\frac{25}{12}\right]\)

\(=\frac{11}{12}-\left(-\frac{5}{12}\right)-\left(-\frac{13}{12}\right)\)

\(=\frac{11}{12}+\frac{5}{12}+\frac{13}{12}\)

\(=\frac{29}{12}\)

\(=\left(1+3+5+...+2019\right)-\left(2+4+...+2018\right)\\ =\dfrac{\left(2019+1\right)\left[\left(2019-1\right):2+1\right]}{2}-\dfrac{\left(2018+2\right)\left[\left(2018-2\right):2+1\right]}{2}\\ =\dfrac{1020100}{2}-\dfrac{1019090}{2}=505\)

giải thích nữa nha

1. Từ 2 đến 100 có 50 số chẵn: (100 - 1 + 1 ) : 2 = 50

  Có 25 cặp số có tổng là 102 = 1 + 100, ....

   Tổng số là : (2+ 100) x25 = 2550.

2. Từ 2  đến 99 có 99 số: 99 - 2 + 1 = 98

 Có 46 cặp số có kq là:  -1 = 2- 3, .....

Vậy kết quả 46. ( -1 ) + 1= - 45

   cứ 3 số liên tiếp nhau tạo thành 1 nhóm :