Bài 2:

a: \(\left(-1\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot\left(-5\right)\cdot\left[\left(-3\right)-\left(-5\right)\right]\)

\(=-\left(1\cdot2\cdot3\cdot4\cdot5\right)\cdot\left[-3+5\right]\)

\(=-120\cdot2=-240\)

b: \(1-2+3-4+5-6+...-98+99\)

=(-1)+(-1)+...+(-1)+99

=99-49=50

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2a) (-1).(-2).(-3).(-4).(-5).[(-3)-(-5)]

= (-1).(-2).(-3).(-4).(-5).2

=> -240

2b) 1-2+3-4+5-6+...+98+99

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

=> Ta có 98 cặp

= (-1)+(-1)+(-1)+...+(-1)+99

= 98(-1)+99

= (-98)+99

= 1

3a) (x-1)(y-2) = 5

=> x-1;y-2 \(\in\) Ư(5) = {-1,-5,1,5}

Ta có bảng :

Vậy x = {0,-4,2,6}

2b) x(y-3)=12

=> x;y-3 \(\in\) Ư(12) = {-1,-2,-3,-4,-12,1,2,4,12}

Tương đương với x = {-1,-2,-3,-4,-12,1,2,4,12}

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)

\(A=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{99}\right)\)

\(A=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)

\(A=\frac{2}{99}\)

A = 2/3 . 3/4 . 4/5 ..... 98/99

A = 2 . 3 . 4 ....... 98/3 . 4 . 5 ...... 99

A = 2/99

A=(1-1/3)x(1-1/4)...x(1-1/99)

  =2/3x3/4x4/5x...x98/99

A = 2/99 nhé !

kết  bạn với mình nha mình hết lượt  rồi

\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)....\left(1-\frac{1}{99}\right)\)

\(=\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{5}{5}-\frac{1}{5}\right)....\left(\frac{99}{99}-\frac{1}{99}\right)\)

\(=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{98}{99}=\frac{2}{99}\)