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

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

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

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

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

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

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

\(\Leftrightarrow\left(x^2-1\right).\left(x^2-3\right)\)

Chúc bạn học tốt !

a) \(x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1\)
Nhóm các hạng tử:

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

Đặt \(t = x^{2}\) thì \(x^{4} - 2 x^{2} + 1 = \left(\right. t - 1 \left.\right)^{2} = \left(\right. x^{2} - 1 \left.\right)^{2}\).
Vậy

\(\boxed{x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1 = \left(\right. x - 1 \left.\right) \left(\right. x^{2} - 1 \left.\right)^{2} = \left(\right. x - 1 \left.\right)^{3} \left(\right. x + 1 \left.\right)^{2} .}\)

b) \(x^{3} - 5 x^{2} - 14 x\)
Lấy \(x\) chung:

\(x \left(\right. x^{2} - 5 x - 14 \left.\right) = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .\)

\(\boxed{x^{3} - 5 x^{2} - 14 x = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .}\)

c) \(2 x^{2} + 2 x y - 4 y^{2}\)
Lấy \(2\) chung: \(2 \left(\right. x^{2} + x y - 2 y^{2} \left.\right)\).
Nhân tử hóa: \(x^{2} + x y - 2 y^{2} = \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right)\).
\(\boxed{2 x^{2} + 2 x y - 4 y^{2} = 2 \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right) .}\)

d) \(3 x^{2} + 8 x y - 3 y^{2}\)
Thử phân tích:

\(3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .\)

\(\boxed{3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .}\)

e) \(x^{2} - x - x y - 2 y^{2} + 2 y\)
Gộp lại theo \(x\): \(x^{2} + x \left(\right. - 1 - y \left.\right) + \left(\right. - 2 y^{2} + 2 y \left.\right)\).
Định thức là một bình phương → nghiệm \(x = 2 y\) và \(x = 1 - y\).
Vậy

\(\boxed{x^{2} - x - x y - 2 y^{2} + 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x + y - 1 \left.\right) .}\)

f) \(x^{2} + 2 y^{2} - 3 x y + x - 2 y\)
Xem như phương trình bậc hai theo \(x\): nghiệm \(x = 2 y\) và \(x = y - 1\).
Do đó

\(\boxed{x^{2} + 2 y^{2} - 3 x y + x - 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x - y + 1 \left.\right) .}\)

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

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

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

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

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

b: \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2-7x+2x-14\right)\)

=x[x(x-7)+2(x-7)]

=x(x-7)(x+2)

c: \(2x^2+2xy-4y^2\)

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

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

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

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

d: \(3x^2+8xy-3y^2\)

\(=3x^2+9xy-xy-3y^2\)

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

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

e: \(x^2-x-xy-2y^2+2y\)

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

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

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

=(x-2y)(x+y-1)

f: \(x^2+2y^2-3xy+x-2y\)

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

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

=(x-2y)(x-y+1)

\(x^4+5x^3+10x-4\)

\(=x^4+5x^3-2x^2+2x^2+10x-4\)

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

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

Mình cũng vừa làm được cách 2:

\(x^4+5x^3+10x-4\)

=\(x^4-4+5x^3+10x\)

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

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

Mình xin lỗi nhé, để mình sửa lại : ^^

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

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

b) \(x^4+5x^2+9=\left(x^4+x^3+3x^2\right)-\left(x^3+x^2+3x\right)+3\left(x^2+x+3\right)\)

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

\(C=x^3+5x^2+8x+4\)

   \(=x^3+x^2+4x^2+4x+4x+4\)

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

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

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

\(D=x^3-x^2-4\)

    \(=x^3-2x^2+x^2-2x+2x-4\)

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

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

Chúc bạn học tốt.

a) \(x^2yz+4zyx+4yz\)

\(=yz\left(x^2+4x+4\right)\)

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

b) \(5x^4-3x^3y-45x^2y^2+27xy^3\)

\(=x\left(5x^3-3x^2y-45xy^2+27y^3\right)\)

v

Ta có : \(4x^2-3x-1\)

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

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

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

Ta có : \(x^2-7x+12\)

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

\(=x\left(x-3\right)-\left(4x-12\right)\)

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

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

a) x² – 4x + 3 = x² – x - 3x + 3

= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

b) x² + 5x + 4 = x² + 4x + x + 4

= x(x + 4) + (x + 4)

= (x + 4)(x + 1)

c) x² – x – 6 = x² +2x – 3x – 6

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

= (x + 2)(x - 3)

d) x⁴+ 4 = x⁴ + 4x² + 4 – 4x²

= (x² + 2)² – (2x)²

= (x² + 2 – 2x)(x² + 2 + 2x)

Bài giải:

a) x² – 4x + 3 = x² – x - 3x + 3

= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

b) x² + 5x + 4 = x² + 4x + x + 4

= x(x + 4) + (x + 4)

= (x + 4)(x + 1)

c) x² – x – 6 = x² +2x – 3x – 6

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

= (x + 2)(x - 3)

d) x⁴+ 4 = x⁴ + 4x² + 4 – 4x²

= (x² + 2)² – (2x)²

= (x² + 2 – 2x)(x² + 2 + 2x)

a)45+x³-5x²-9x

(45-9x)+(x³-5x²)

9(5-x)+x²(x-5)

(9-x²)(5-x)

(3-x)(3+x)(5-x)

b)x⁴-2x³-2x²-2x-3

x³(x-2)-2x(x-2)-3

(x-2)(x³-2x)-3

x

a) \(4x^4-21x^2y^2+y^4\)

Ấn nhầm :v

a) \(4x^4-21x^2y^2+y^4\)

\(=\left(2x^2\right)^2-2\cdot2x^2\cdot y^2+y^2-25x^2y^2\)

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

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

b) \(x^5-5x^3+4x\)

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

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

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

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

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