14.D

15.B

16.C

d b c

II. 

1B

2A

3C

4B

5D

6B

7D

8D

Bài 5:

a) Do \(x,y\in N\)

\(\Rightarrow\left\{\left(x;y-2\right)\right\}\in\left\{\left(1;7\right),\left(7;1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;9\right),\left(7;3\right)\right\}\)

b) Do \(x,y\in N\)

\(\Rightarrow\left(x+1;y+5\right)\in\left\{\left(1;12\right),\left(2;6\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;7\right),\left(1;1\right)\right\}\)

c) Do \(x,y\in N\)

\(\Rightarrow\left(x-1;2y+1\right)\in\left\{\left(18;1\right),\left(2;9\right),\left(6;3\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(19;0\right),\left(3;4\right),\left(7;1\right)\right\}\)

\(5,\\ a,=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\\ b,=x^4+16x^2+64-16x^2=\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\\ c,=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+1\\ =x^6\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^6-x^4+x^3-x+1\right)\left(x^2+x+1\right)\)

\(d,=x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ =\left(x^4-x^2+1\right)\left(x^4+2x^2+1-x^2\right)\\ =\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\\ e,=x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1\\ =x^3\left(x^2+x+1\right)-x^2\left(x^2+x+x\right)+\left(x^2+x+1\right)\\ =\left(x^3-x^2+1\right)\left(x^2+x+1\right)\\ f,=x^3+2x^2-x^2-2x+2x+4\\ =\left(x+2\right)\left(x^2-x+2\right)\\ g,=x^4+2x^2+1-25=\left(x^2+1\right)^2-25\\ =\left(x^2+1-5\right)\left(x^2-1-5\right)=\left(x^2-4\right)\left(x^2-6\right)=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)

\(h,=x^3-2x^2+2x^2-4x+2x-4=\left(x-2\right)\left(x^2+2x+2\right)\\ i,=a^4-4a^2b^2+4b^4-4a^2b^2=\left(a^2-2b^2\right)^2-4a^2b^2\\ =\left(a^2-2ab-2b^2\right)\left(a^2+2ab-2b^2\right)\)

Lời giải:
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:

$|x+2020|+|x+2021|=|x+2020|+|-(x+2021)|$

$\geq |x+2020-(x+2021)|=1$

Vậy GTNN của biểu thức là $1$. Giá trị này đạt tại $(x+2020).-(x+2021)\geq 0$

$(x+2020)(x+2021)\leq 0$

$-2021\leq x\leq -2020$

a) Đặt \(a=x^2+x\)

Đa thức trở thành: \(a^2-14a+24=\left(a^2-14a+49\right)-25=\left(a-7\right)^2-25=\left(a-7-5\right)\left(a-7+5\right)=\left(a-12\right)\left(a-2\right)\)

Thay a:

\(\left(a-12\right)\left(a-2\right)=\left(x^2+x-12\right)\left(x^2+x-2\right)\)

b) Đặt \(a=x^2+x\)

Đa thức trở thành:

\(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)-12=a^2+4a-12=\left(a^2+4x+4\right)-16=\left(a+2\right)^2-16=\left(a+2-4\right)\left(a+2+4\right)=\left(a-2\right)\left(a+6\right)\)

Thay a:

\(\left(a-2\right)\left(a+6\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)\)

Em nên chia bài này ra làm 2, 3 nhé vì liền một lúc 4 đoạn như vậy mng sẽ nản hoặc khó để làm em nhé!