Bài này trên mạng cũng có mà.

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

\(=x\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^3-16x-x^4+1\)

bạn ktra lại đề

b)  \(x^4+2x^3+5x^2+4x-12\)

\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

Ủa pạn có thể giải ại cái bước thứ 2 đc ko ạk

\(=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

\(x^4+4x^2-5\)

\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-9\)

\(=\left(x^2+2\right)^2-9\)

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

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

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

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

a)  \(x^4+4x^2-5=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9\)

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

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

b)  \(x^6-x^4+2x^3+2x^2=x^4\left(x^2-1\right)+2x^2\left(x+1\right)\)

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

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

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

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

a) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ac\)

\(=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2ac\)

\(=b^2-2bc+2ac=b.\left(b-2c+2a\right)\)

b) \(x^4+2x^3+5x^2+4x-12\)

\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)

\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

\(=\left(x-1\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\)

\(=\left(x-1\right)\left[x^2.\left(x+2\right)+x.\left(x+2\right)+6.\left(x+2\right)\right]\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

Pạn Khánh Châu ơi

Cái dòng thứ 2 đấy, dấu hiệu nhận biết là j vậy

Mà sao pạn phân tích hay vậy????

\(x^5+x^4+1\)

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

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

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

Mình làm giống ban kia nha

k tui nha

thanks

 Câu trả lời hay nhất:  x7 + x2 + 1 = (x7 – x) + (x2 + x + 1) 
= x.(x6 – 1) + (x2 + x +1) 
= x.(x3 - 1).(x3 +1) + (x2 + x +1) 
= x.(x-1).(x2 + x +1).(x3 +1) + (x2 + x +1) 
= (x2 + x +1).[x.(x-1).(x3 +1) + 1] 
= (x2 + x +1).[(x2-x).(x3 +1) + 1] 
= (x2 + x +1).(x5-x4 + x2 -x + 1)

1/ = x⁴ + 2x³ + 4x² + 3x - 10 = (x⁴ - x³) + (3x³ - 3x²) + (7x² - 7x) + (10x - 10)

= (x - 1)(x³ + 3x² + 7x + 10) = (x - 1)[(x³ + 2x²) + (x² + 2x) + (5x + 10)]

= (x - 1)(x + 2)(x² + x + 5)

2/ = (x⁵ - 2x⁴) + (x⁴ - 2x³) + (x³ - 2x²) + (x² - 2x) + (x - 2) = (x - 2)(x⁴ + x³ + x² + x + 1)

\(a,\)

\(x^3y-y\)

\(=y\left(x^3-1\right)\)

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

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

\(b,\)

\(x^3y+y\)

\(=y\left(x^3+1\right)\)

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

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

\(c,\)

\(\left(x-y\right)^2-x\left(y-x\right)\)

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

\(=\left(x-y\right)\left(x-y+x\right)\)

\(=\left(x-y\right)\left(2x-y\right)\)