a) \(x^2+7x+12\)

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

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

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

b) \(3x^2-5x+2\)

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

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

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

a) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 4)(x + 3)

b) 3x² - 5x + 2 = 3x² - 3x - 2x + 2 = 3x(x - 1) - 2(x - 1) = (3x - 2)(x - 1)

c) x² + 9x - 10 = x² + 10x - x - 10 = x(x + 10) - (x + 10) = (x - 1)(x + 10)

d) x² - 7x - 8 = x² - 8x + x - 8 = x(x - 8) + (x - 8) = (x + 1)(x - 8)

e) 2x² + 3x - 5 = 2x² + 5x - 2x - 5 = x(2x + 5) - (2x + 5) = (x  - 1)(2x + 5)

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Nháp:

Ta nhẩm nghiệm được \(a=-3\) nên khi phân tích nó sẽ có nhân tử là \(x+3\)

Giả sử khi phân tích thành nhân tử nó sẽ có dạng:\(\left(x+3\right)\left(x^3+ax^2+bx+c\right)\)

\(=x^4+ax^3+bx^2+cx+3x^3+3ax^2+3bx+3c\)

\(=x^4+\left(a+3\right)x^3+\left(3a+b\right)x^2+\left(c+3b\right)x+3c\)

Mà \(\left(x+3\right)\left(x^3+ax^2+bx+c\right)=x^4+4x^3+5x^2+7x+3\)

Cân bằng hệ số ta được:

\(a=1;b=2;c=1\)

Khi đó \(x^4+4x^3+5x^2+7x+3=\left(x+3\right)\left(x^3+x^2+2x+1\right)\)

Bài làm

Ta có:

\(x^4+4x^3+5x^2+7x+3\)

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

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

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

P/S:Mik nghĩ đến đây là hết rồi:3

a) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

Đặt \(x^2+x=t\), đa thức trở thành : \(t^2-2t-15\)

= \(\left(t+3\right)\left(t-5\right)\)

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

b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+2ab+2ac+2bc-a^3-b^3-c^3\)

\(=2ab+2ac+2bc=2\left(ab+ac+bc\right)\)

c) \(x-1+x^{n+3}-x^n\)

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

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

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

d) \(2x^4-7x^3-2x^2+13x+6\)

\(=\left(2x^4+2x^3\right)-\left(9x^3+9x^2\right)+\left(7x^2+7x\right)+\left(6x+6\right)\)

\(=\left(x+1\right)\left(2x^3-9x^2+7x+6\right)\)

\(=\left(x+1\right)\left[\left(2x^3+x^2\right)-\left(10x^2+5x\right)+\left(12x+6\right)\right]\)

\(=\left(x+1\right)\left(2x+1\right)\left(x^2-5x+6\right)\)

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

Chữa lại câu b :3

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

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

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

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

b) \(3x^2-7x-10=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

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

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

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

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

b) \(3x^2-7x-10=\left(3x^2+3x\right)-\left(10x+10\right)\)

= \(3x\left(x+1\right)-10\left(x+1\right)\)

= \(\left(x+1\right)\left(3x-10\right)\)

a) Ta có: \(8x^2+30x+7\)

\(=8x^2+28x+2x+7\)

\(=4x\left(2x+7\right)+\left(2x+7\right)\)

\(=\left(2x+7\right)\left(4x+1\right)\)

b) Ta có: \(4x^3-12x^2+9x\)

\(=x\left(4x^2-12x+9\right)\)

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

c) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

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

\(=\left(x+2\right)\cdot3x\)

d) Ta có: \(ab+c^2-ac-bc\)

\(=\left(ab-bc\right)+\left(c^2-ac\right)\)

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

\(=b\left(a-c\right)-c\left(a-c\right)\)

\(=\left(a-c\right)\left(b-c\right)\)

e) Ta có: \(4x^2-y^2+1-4x\)

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

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

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

f) Ta có: \(6x^2-7x-20\)

\(=6x^2-15x+8x-20\)

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

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

\(4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\), \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)

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

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

\(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(2x-5\right)\left(3x+4\right)\)

\(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(4x+1\right)\left(2x+7\right)\)

Bài 1:

a) \(x^2-2xy-25+y^2\) (Sửa đề)

\(=x^2-2xy+y^2-25\)

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

\(=\left(x-y-5\right)\left(x-y+5\right)\)

Vậy ...

b) \(x\left(x-1\right)+y\left(1-x\right)\)

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

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

Vậy ...

c) \(7x+7y-\left(x+y\right)\) (Sửa đề)

\(=7\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(7-1\right)\)

\(=6\left(x+y\right)\)

Vậy ...

d) \(x^4+y^4\)

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

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

\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)

Vậy ...

Bạn xem lại 1 sô câu sai và bài 2 hộ mk

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

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

\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)