Bài 2: 

a: M=12-x+x-73+96+x-23=x+12=113

a)12/23

   18/39

    24/52

   30/65 

   36/78

   42/91

b) 45/-21

    75/-35

a: (326-43)+(174-57)

=326-43+174-57

=500-100

=400

b: (351-875)-(125-149)

=351-875-125+149

=500-1000

=-500

c: \(-415-\left\lbrace-215-\left\lbrack-115-\left(-315-2016\right)\right\rbrack\right\rbrace\)

=-415+215+[-115+315+2016]

=-200+[200+2016]

=-200+200+2016

=2016

d: [461+(-78)+40]+(-461)

=461-78+40-461

=-78+40

=-38

e: -323+[(-874)+564-241]
=-323-241+564-874

=-564+564-874

=-874

f: [(-59)+71]-[-83-(-95)]

=-59+71-(-83+95)

=12-12

=0

a: (a-b)+(c-d)-(a+c)

=a-b+c-d-a-c

=-b-d=-(b+d)

b: (a-b)-(c-d)+(b+c)

=a-b-c+d+b+c

=a+d

c: (a+b-c)+(a-b)-(a-b-c)

=a+b-c+a-b-a+b+c

=a+b

d: -(a-b-c)+(-a+b-c)-(-a+b+c)

=-a+b+c-a+b-c+a-b-c

=-a+b-c=-(a-b+c)

Số số hạng của dãy số là:

\(\frac{999-1}{2}+1=\frac{998}{2}+1=\frac{1000}{2}=500\) (số)

-1-3+5+7-9-11+13+15-...-993-995+997+999

=(-1-3+5+7)+(-9-11+13+15)+...+(-993-995+997+999)

=8+8+...+8

\(=8\cdot\frac{500}{4}=500\cdot2=1000\)

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

x+1;y-2 thuộc Ư(3){1;-1;3;-3}

ta có bảng sau :

vậy cặp x;y thuộc {(2;3);(0;1);(4;5);(-2;-1)}
 

Bài 4:

a: (x+1)(y-2)=3

=>(x+1;y-2)∈{(1;3);(3;1);(-1;-3);(-3;-1)}

=>(x;y)∈{(0;5);(2;3);(-2;-1);(-4;1)}

b: (2x-1)(3y+1)=12

mà 2x-1 lẻ

nên (2x-1;3y+1)∈{(1;12);(-1;-12);(3;4);(-3;-4)}

=>(2x;3y)∈{(2;11);(0;-13);(4;3);(-2;-5)}

mà 3y⋮3

nên (2x;3y)∈{(4;3)}

=>(x;y)∈{(2;1)}

c: xy-x+y-2=0

=>x(y-1)+y-1-1=0

=>(x+1)(y-1)=1

=>(x+1;y-1)∈{(1;1);(-1;-1)}

=>(x;y)∈{(0;2);(-2;0)}

Bài 3:

a: 3n-1⋮n+2

=>3n+6-7⋮n+2

=>-7⋮n+2

=>n+2∈{1;-1;7;-7}

=>n∈{-1;-3;5;-9}

b:2n+3⋮3n-1

=>6n+9⋮3n-1

=>6n-2+11⋮3n-1

=>11⋮3n-1

=>3n-1∈{1;-1;11;-11}

=>3n∈{2;0;12;-10}

=>n∈{2/3;0;4;-10/3}

mà n là số nguyên

nên n∈{0;4}

Đặt A=\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)

         =(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\))+(\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\))+...+\(\dfrac{1}{70}\)

Nhận xét:

\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\)>\(\dfrac{1}{20}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{1}{20}\)=\(\dfrac{10}{20}\)=\(\dfrac{1}{2}\)

\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\)>\(\dfrac{1}{30}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{30}\)=\(\dfrac{10}{30}\)=\(\dfrac{1}{3}\)

\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\)>\(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+...+\(\dfrac{1}{60}\)=\(\dfrac{30}{60}\)=\(\dfrac{1}{2}\)

=>A>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)+\(\dfrac{1}{61}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)=\(\dfrac{4}{3}\)

=>A>\(\dfrac{4}{3}\)

Vậy: \(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{4}{3}\) (ĐPCM)

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Bài 3: 

b: \(x\in\left\{3;-3\right\}\)

Tôi không biết là ai cũng muốn em nó 3