1: Ta có: \(A=25x^4-24x^2-1\)

\(=25x^4-25x^2+x^2-1\)

\(=\left(x^2-1\right)\left(25x^2+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(25x^2+1\right)\)

2: Ta có: \(A=64x^4+63x^2-1\)

\(=64x^4+64x^2-x^2-1\)

\(=\left(x^2+1\right)\left(64x^2-1\right)\)

\(=\left(x^2+1\right)\left(8x-1\right)\left(8x+1\right)\)

3: Ta có: \(A=x^4-15x^2+50\)

\(=x^4-5x^2-10x^2+50\)

\(=\left(x^2-5\right)\left(x^2-10\right)\)

4: Ta có: \(A=-10x^4+9x^2+1\)

\(=-10x^4+10x^2-x^2+1\)

\(=\left(x^2-1\right)\left(-10x^2-1\right)\)

\(=-\left(10x^2+1\right)\left(x-1\right)\left(x+1\right)\)

7. A = (x + y)^2 - 4y^2

= (x + y - 2y)(x + y + 2y)

= (x - y)(x + 3y)

2. x^4 + 4

= x^4 + 4x^2 + 4 - 4x^2

= (x^2 + 2)^2 - (2x)^2

= (x^2 + 2x + 2)(x^2 - 2x + 2)

3. 4x^4 + 16

= 4(x^4 + 4)

= 4(x^2 + 2x + 2)(x^2 - 2x + 2)

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 $(x+2020).-(x+2021)\geq 0$

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

$-2021\leq x\leq -2020$

Câu 3: 

Số cần tìm là 

(99-74):5=25:5=5

37: Ta có: \(A=8x^2-2x-3\)

\(=8x^2-6x+4x-3\)

\(=2x\left(4x-3\right)+\left(4x-3\right)\)

\(=\left(4x-3\right)\left(2x+1\right)\)

38: Ta có: \(A=8x^2+2x-3\)

\(=8x^2+6x-4x-3\)

\(=2x\left(4x+3\right)-\left(4x+3\right)\)

\(=\left(4x+3\right)\left(2x-1\right)\)

39: Ta có: \(A=-8x^2+5x+3\)

\(=-8x^2+8x-3x+3\)

\(=-8x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-1\right)\left(-8x-3\right)\)

40: Ta có: \(A=8x^2-10x-3\)

\(=8x^2-12x+2x-3\)

\(=4x\left(2x-3\right)+\left(2x-3\right)\)

\(=\left(2x-3\right)\left(4x+1\right)\)

41: Ta có: \(A=8x^2+10x-3\)

\(=8x^2+12x-2x-3\)

\(=4x\left(2x+3\right)-\left(2x+3\right)\)

\(=\left(2x+3\right)\left(4x-1\right)\)

42: Ta có: \(A=-8x^2+23x+3\)

\(=-8x^2+24x-x+3\)

\(=-8x\left(x-3\right)-\left(x-3\right)\)

\(=\left(x-3\right)\left(-8x-1\right)\)

43: Ta có: \(A=8x^2-23x-3\)

\(=8x^2-24x+x-3\)

\(=8x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(8x+1\right)\)

44: Ta có: \(A=10x^2-11x-6\)

\(=10x^2-15x+4x-6\)

\(=5x\left(2x-3\right)+2\left(2x-3\right)\)

\(=\left(2x-3\right)\left(5x+2\right)\)

45: Ta có: \(A=-10x^2+11x+6\)

\(=-10x^2+15x-4x+6\)

\(=-5x\left(2x-3\right)-2\left(2x-3\right)\)

\(=\left(2x-3\right)\left(-5x-2\right)\)

\(A=x^7-4x^3+x^2+2=x^3\left(x^4-4\right)+x^2+2\)

\(=x^3\left(x^2-2\right)\left(x^2+2\right)+x^2+2\)

\(=\left(x^2+2\right)\left(x^3\left(x^2-2\right)+1\right)\)

\(=\left(x^2+2\right)\left(x^5-2x^3+1\right)\)

\(=\left(x^2+2\right)\left(x^5-x^4+x^4-x^3-x^3+x^2-x^2+x-x+1\right)\)

\(=\left(x^2+2\right)\left[x^4\left(x-1\right)+x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\right]\)

\(=\left(x^2+2\right)\left(x-1\right)\left(x^4+x^3-x^2-x-1\right)\)