b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

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

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

c: \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)

d: \(4x^2+12x+9-y^2\)

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

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

Các phần về đa thức bạn có thể vào đây giải trực tuyến luôn bạn nhé.

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a,2x²-7x+6=(2x²-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x²+x-6=(x²+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x³+3x²+6x+4=x³+x²+2x²+2x+4x+4
=(x+1)(x²+2x+4)
d,x¹⁰+x⁵+1=(x¹⁰-x)+(x⁵-x²)+(x²+x+1)
=x((x³)³-1)+x²(x³-1)+(x²+x+1)
=x(x³-1)(x⁶+x³+1)+x²(x-1)(x²+x+1)+(x²+x+1)
=x(x-1)(x²+x+1)+x²(x-1)(x²+x+1)+(x²+x+1)
(x²+x+1)(x²-x+x³-x²+1)
e,(12x²-12xy+3y²)-10x(2x-y)=3(4x²-4xy+y²)-10x(2x-y)
=3(2x-y)²-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)

phân tích đa thức thành nhân tử

a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)

\(2x^2-7x+6\)

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

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

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

 a) 3x2 – 7x + 2

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

\(=\left(3x^2-6x\right)-\left(x-2\right)\)

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

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

 b) a(x2 + 1) – x(a2 + 1)

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

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

.......?

a) Ta có: \(3x^2-7x+2\)

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

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

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

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

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

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))

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

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

d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

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

b: \(3x^2+2x-5\)

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

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

c: \(3-2x-x^2\)

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

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

d: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

a, x^2 + 5x +4

= x^2 + 1x + 4x + 4

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

= x ( x + 1 ) + 4 ( x + 1 )

= (x + 1) (x + 4)

b, x^2 - 6x + 5

= x^2 - 1x - 5x + 5

= (x^2 - 1x) - (5x - 5)

= x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

c, x^2 + 7x + 12

= x^2 + 3x + 4x + 12 

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

= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

d, 2x^2 - 5x + 3

= 2^x2 - 2x - 3x + 3

= 2x (x - 1) - 3 (x - 1)

= (x-1) (2x - 3)

e, 7x  - 3x^2 - 4

= 3x + 4x - 3x^2 - 4

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

= 3x (1 - x) + 4 (x - 1)

= 3x (1-x) - 4 (1 - x)

= (1 - x) (3x - 4)

f, x^2 - 10x + 16

= x^2 - 2x - 8x + 16

= (x^2 - 2x) - (8x - 16)

= x (x - 2) - 8 (x - 2)

= (x - 2) (x - 8)

a, (x+1)(x+4)

b,(x-5)(x-1)

c,(x+3)(x+4)

d,(2x-3)(x-1)

e,(-3x+4)(x-1)

f, (x-8)(x-2)

C

x^10 + x^5 + 1 
= x^10 + x^9 - x^9 + x^8 - x^8 + x^7 - x^7 + x^6 - x^6 + x^5 + x^5 - x^5 + x^4 - x^4 + x^3 - x^3 + x^2 - x^2 + x - x + 1 
= (x^10 + x^9 + x^8) - (x^9 + x^8 + x^7) + (x^7 + x^6 + x^5) - (x^6 + x^5 + x^4) + (x^5 + x^4 + x^3) - (x^3 + x^2 + x) + (x^2 + x + 1) 
= x^8 (x^2 + x + 1) - x^7 (x^2 + x + 1) + x^5 (x^2 + x + 1) - x^4 (x^2 + x + 1) + x^3 (x^2 + x + 1) - x (x^2 + x + 1) + (x^2 + x + 1) 
= (x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) 
----------------------- 
Phương pháp: 
Khi gặp bài toán phân tích thành nhân tử dạng x^(3m + 1) + x^(3n + 2) + 1 em thêm bớt các hạng tử từ bậc cao nhất trừ đi 1 đến x (bậc nhất) sao cho tổng số các hạng tử trong đa thức mới là một bội của 3. Sau đó nhóm ba hạng tử một sao cho trong mỗi nhóm có x² + x + 1 
Dạng này khi phân tích luôn có kết quả là: (x² + x + 1).Q(x)

x^7 + x^2 + 1 = x^7 + x^6 - x^6 + x^5 - x^5 + x^4 - x^4 +x^3 - x^3 +2x^2 - x^2 +x - x +1 
=(x^7 + x^6 + x^5) - (x^6 +x^5 +x^4) + (x^4 + x^3 +x^2) - (x^3 +x^2 + x) + (x^2 + x +1) 
=x^5(x^2 + x + 1) - x^4(x^2 + x + 1) +x^2(x^2 + x + 1) - x(x^2 + x + 1) + (x^2 + x + 1) 
=(x^2 + x + 1)(x^5 - x^4 +x^2 -x +1)