1. b3+b= 3                                       

(b3+b)=3                            

b.(3+1)=3

b. 4= 3

b=\(\dfrac{3}{4}\)

a3+a= 3                                       b3

(a3+a)=3                            

a.(3+1)=3

a. 4= 3

a=\(\dfrac{3}{4}\)

2

Hiển nhiên \(c\left(c+1\right)>a\left(a+1\right)\Rightarrow c>a\ge b\)

Nếu \(c\ge2a\Rightarrow c\left(c+1\right)\ge2a\left(2a+1\right)=4a^2+2a\)

Mà \(a\left(a+1\right)+b\left(b-1\right)\le a\left(a+1\right)+a\left(a-1\right)=2a^2\)

\(\Rightarrow2a^2\ge4a^2+2a\Rightarrow2a^2+2a\le0\) (vô lý)

\(\Rightarrow c< 2a\)

Ta có:

\(4a\left(a+1\right)+4b\left(b-1\right)+1=4c\left(c+1\right)+1\)

\(\Leftrightarrow4a\left(a+1\right)+\left(2b-1\right)^2=\left(2c+1\right)^2\)

\(\Leftrightarrow4a\left(a+1\right)=\left(2c+1\right)^2-\left(2b-1\right)^2\)

\(\Leftrightarrow a\left(a+1\right)=\left(c-b+1\right)\left(c+b\right)\) (*)

Nếu \(c-b+1\ge a\Rightarrow\left(c-b+1\right)\left(c+b\right)>a\left(a+b\right)>a\left(a+1\right)\) (ktm)

\(\Rightarrow c-b+1< a\) \(\Rightarrow c-b+1\) ko có ước nguyên tố nào là a

\(\Rightarrow c+b⋮a\Rightarrow\dfrac{c+b}{a}\in Z\) (1)

Theo chứng minh ban đầu, ta có \(b\le a< c< 2a\)

\(\Rightarrow a< c+b< 2a+a=3a\Rightarrow1< \dfrac{c+b}{a}< 3\) (2)

(1);(2) \(\Rightarrow\dfrac{c+b}{a}=2\Rightarrow c+b=2a\)

Thế vào (*) \(\Rightarrow a+1=2\left(c-b+1\right)\Rightarrow2c-2b+1=a\)

\(\Rightarrow2\left(2a-b\right)-2b+1=a\Rightarrow3a-4b+1=0\)

\(\Rightarrow3\left(a-1\right)=4\left(b-1\right)\)

\(\Rightarrow b-1⋮3\Rightarrow b-1=3k\Rightarrow b=3k+1\)

\(\Rightarrow a=4k+1\)

\(\Rightarrow c=2a-b=5k+1\)

\(\Rightarrow A=3\left(5k+1\right)-5\left(3k+1\right)=-2\)

Bài 2:

a: \(x^4-x^3+6x^2-x+a\)

\(=x^4-x^3+5x^2+x^2-x+5+a-5\)

\(=\left(x^2-x+5\right)\left(x^2+1\right)+a-5\)

TA có: \(x^4-x^3+6x^2-x+a\) ⋮\(x^2-x+5\)

=>a-5=0

=>a=5

b: \(x^4+a\cdot x^2+b\)

\(=x^4-x^3+x^2+x^3-x^2+x+a\cdot x^2-a\cdot x+a\) +x(a-1)+b-a

\(=\left(x^2-x+1\right)\left(x^2-x+a\right)\) +x(a-1)+b-a

Để \(x^4+a\cdot x^2+b\) ⋮\(x^2-x+1\) thì a-1=0 và b-a=0

=>a=b=1

a: \(\Leftrightarrow\left(x;y-3\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;20\right);\left(17;4\right);\left(-1;-14\right);\left(-17;2\right)\right\}\)

b: \(\Leftrightarrow\left(x-1;y+2\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;5\right);\left(8;-1\right);\left(0;-9\right);\left(-6;-3\right)\right\}\)

c: =>(y+1)(3x+1)=7

=>\(\left(3x+1;y+1\right)\in\left\{\left(1;7\right);\left(7;1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;6\right);\left(2;0\right)\right\}\)

\(a.-7< x< -1\\ x\in\left\{-6;-5;-4;-3;-2\right\}\\ \Rightarrow\left(-6\right)+\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)\\ =-20\)

\(b.-1\le x\le6\\ x\in\left\{-1;0;1;2;3;4;5;6\right\}\\ \Rightarrow\left(-1\right)+0+1+2+3+4+5+6\\ =20\)

\(c.-5\le x< 6\\ x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\\ \Rightarrow-5-4+-3+-2+-1+0+1+2+3+4+5\\ =0\)