Phân tích đa thức (x^2+8x+7)(x+3)(x+5) + 15 thành nhân tử

a)\(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)

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

Đặt \(t=x^2+8x+7\) ta có:

\(=t\left(t+8\right)+15=t^2+8t+15\)

\(=\left(t+3\right)\left(t+5\right)\)\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)

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

b)\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

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

Đặt \(t=x^2+7x+10\) ta có:

\(=t\left(t+2\right)-24=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

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

phân tích các đa thức sau thành nhân tử

Phân yichs các đa thức sau thành nhân tử

8) 4x^4-32x^2 + 1

9) 3(x4 + x2 + 1) - (x2 + x + 1)2

10) 64x4 + y4
11) a6 + a4 + a2b2 + b4 - b6
12) x3 + 3xy + y3 - 1
13) 4x4 + 4x3 + 5x2 + 2x + 1
14) x8 + x + 1
15) x8 + 3x4 + 4
16) 3x2 + 22xy + 11x + 37y + 7y2 +10
17) x4 - 8x + 63

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

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

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

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

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

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

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

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

= \(\left(x-1\right)^2.2.\left(x^2+x+1\right)\)

10) \(64x^4+y^4\)

= \(\left(8x^2\right)^2+2.8x^2.y^2+\left(y^2\right)^2-16x^2y^2\)

= \(\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

= \(\left(8x^2+y^2-4xy\right).\left(8x^2+y^2+4xy\right)\)

Bài 1 : Tính hợp lí

a, 413(413-26)+169

b, x²+2x+1 tại x=99

Bài 2: Pân ích các đa thức sau thành nhân tử

a, 9x²-36

b, ax-ay-bx+by

c, y³-4y²+3y

( x² - x - 2)² + ( x - 2)²

= ( x² + x - 2x - 2)² + ( x - 2)²

= [ x( x + 1) - 2( x + 1) ]² + ( x - 2)²

= ( x + 1)²( x - 2)² + ( x - 2)²

= ( x - 2)²[ ( x + 1)² + 1]

= ( x - 2)²( x² + 2x + 2) ( không chắc lắm # ^ . ^ #)

Bài 2:

\(a,9x^2-36\)

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

\(=\left(3x-6\right)\left(3x+6\right)\)

\(b,ax-ay-bx+by\)

\(=\left(ax-ay\right)-\left(bx-by\right)\)

\(=a\left(x-y\right)-b\left(x-y\right)\)

\(=\left(x-y\right)\left(a-b\right)\)

\(c,y^3-4y^2+3y\)

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

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

\(=y\left[\left(y^2-3y\right)-\left(y-3\right)\right]\)

\(=y\left[y\left(y-3\right)-\left(y-3\right)\right]\)

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

phân tích đa thức sau thành nhân tử

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

Phân tích thành nhân tử

3x^3-12x

a).

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

b).

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

đặt: \(t=x^2+3x+1\) khi đó:

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

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

1.

= 4x\(^{^{ }2}\)-4x-9x+9

=4x(x-1)-9(x-1)

=(4x-9)(x-1)

phân tích đa thức sau thành nhân tử:

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

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

Phân tích đa thức thành nhân tử
1, 4x² - 13x + 9
2, 5x² + 17x + 12
3, x² - 3x - 10
4, x² + x - 12
5, x³ + 5x² + 3x - 9
6, x_³ - 7x + 6
7, x⁵ - 5x³ + 4x
8, 4x⁴ - 21x²y² + y⁴
9, 2a²b + 4ab² - a²c + ac² - 4b²c + 2bc² - 4abc
10, 8x³(y+z) - y³(z+2x) - z³(2x-y)

Phân tích đa thức thành nhân tử:

a. a(b\(^2\)+c\(^2\))+b(c\(^2\)+a\(^2\))+c(a\(^2\)+b\(^2\))+2abc

b. a(b-c)\(^3\)+b(c-a)\(^3\)+c(a-b)\(^3\)

c. a\(^2\)b\(^2\)(a-b)+b\(^2\)c\(^2\)(b-c)+c\(^2\)a\(^2\)(c-a)

d. a\(^4\)(b-c)+b\(^4\)(c-a)+c\(^4\)(a-b)

phân tích đa thức thành nhân tử

=a, a(b²+c²)+b(a²+c²)+c(a²+b²)+2abc

= ab²+ac²+ba²+bc²+ca²+cb²+2abc

= c²(a+b)+ab(a+b)+c(a²+b²+2ab)

= c²(a+b)+ab(a+b)+c(a+b)²

= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)

= (a+b)(c²+ab+ca+cb)

= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)

=(a+b)(a+c)(b+c)

b, a(b-c)³+b(c-a)³+c(a-b)³

= a(b-c)³-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)²(a-b)-3b(b-c)(a-b)²-b(a-b)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)(a-b)(b-c+a-b)-b(a-b)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)(a-b)(a-c)-b(a-b)³+c(a-b)³

= (b-c)³(a-b)-3b(b-c)(a-b)(a-c)-(a-b)³(b-c)

= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)

=(b-c)(a-b)(b²-2bc+c²-3ab+3bc-a²+2ab-b²)

= (b-c)(a-b)(c²-a²+bc-ab)

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

= (b-c)(a-b)(c-a)(c+a+b)

c, a²b²(a-b)+b²c²(b-c)+c²a²(c-a)

= a²b²(a-b)-b²c²\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c²a²(c-a)

= a²b²(a-b)-b²c²(a-b)-b²c²(c-a)+c²a²(c-a)

= b²(a-b)(a²-c²)+c²(c-a)(a²-b²)

= b²(a-b)(a-c)(a+c)-c²(a-c)(a-b)(a+b)

= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)

= (a-c)(a-b)(b²a+b²c-c²a-c²b)

= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)

= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)

= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)

= (a-c)(a-b)(b-c)(ab+ac+bc)

d, a⁴(b-c)+b⁴(c-a)+c⁴(a-b)

= a⁴(b-c)-b⁴[(b-c)+(a-b)]+c⁴(a-b)
= (b-c)(a⁴-b⁴)+(a-b)(c⁴-b⁴)
= (b-c)(a²-b²)(a²+b²)+(a-b)(c²-b²)(c²+b²)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b²+c²)
= (b-c)(a-b)(a³+ab²+ba²+b³-bc²-b³-cb²-c³)

= (b-c)(a-b)(a³+ab²+ba²-bc²-c³-cb²)
= (b-c)(a-b)(a³-c³)+b²(a-c)+b(a²-c²)
= (b-c)(a-b)(a-c)(a²+ac+c²)+b²(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a²+ac+c²+b²+ab+ac)

= (a-b)(b-c)(c-a)(a²+b²+c²+ab+bc+ca)

\(3x^3-12x\)

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

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

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