Phân tích đa thức x^5+x+1 thành nhân tử

a).

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

b).\(x^8+x^7+1=\left(x^8+x^7+x^6\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

d).

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

e).

\(x^8+x^4+1=x^8+2x^4+1-x^4\\ =\left(x^4+1\right)^2-\left(x^2\right)^2\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\\ =\left(x^4-x^2+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)

Phân tích đa thức thành nhân tử

a, 5x-10\(x^2\)

b, \(\dfrac{1}{2}x\left(x^2-4\right)+4\left(y+2\right)\)

c, \(x^4-y^6\)

d, \(x^3+y\left(1-3x^2\right)+z\left(3y^2-1\right)-y^3\)

e, \(x^3-4x^2+4x-1\)

f, \(x^2+2xy-8y^2+2xz+14yz+3z^2\)

g, \(x^4+6x^3-12x^2-8x\)

h, \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

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

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

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

g) \(x^4+6x^3-12x^2-8x\)

= \(x\left(x^3-2x^2+8x^2-16x+4x-8\right)\)

= \(x\left[x^2\left(x-2\right)+8x\left(x-2\right)+4\left(x-2\right)\right]\)

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

h) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (*)

Đặt \(x^2+4x+8=a\) => (*) trở thành:

\(a^2+3ax+2x^2\) = \(a^2+ãx+2ax+x^2\)

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

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

Thay \(a=x^2+4x+8\) vào (1) ta được:

\(\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

=\(\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)\)

= \(\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)

= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

P/s: Còn câu f đang suy nghĩ!

phân tích các đa thức sau thành nhân tử

a) \(27x^3-27x^2+18x-4\)

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

Ai giúp mk nha !!!

  bởi thanh hằng 17/10/2018

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a, 27x³ - 27x² +18x-4=27x³ - 9x² -18x² +6x +12x-4

= 9x²(3x-1) - 6x(3x-1) + 4(3x-1)

= (3x-1)(9x²-6x+4)

b, 2x³-x²+5x+3 =\(2x^3+x^2-2x^2-x+6x+3\)

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

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

Phân tích các đa thức sau thành nhân tử

1) a) \(x^3-9x^2+6x+16\)

b) \(x^3-x^2-x-2\)

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

d) \(x^3-6x^2-x+30\)

2) \(x^3-7x-6\) (giải bằng nhiều cách)

Nghĩ mãi mà ko ra,ai giúp mk đi !!!

  bởi Nguyễn Quang Minh Tú 18/10/2018

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Bài 2:

Cách 1:

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

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

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

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

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

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

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

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

Cách 2:

\(x^3-7x-6=x^3+x^2-x^2-x-6x-6\)

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

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

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

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

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

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

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

Chúc bạn học tốt!!! Còn 1 cách nữa nhưng mình mỏi tay quá!!!

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

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

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

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

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

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

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

phân tích đa thức thành nhân tử

a^{2(b-c)+b2(c-a)+c2(a-b)}

X⁴ - 4x³ - 7x² +16x -3

Dùng pp hệ số bất định nha bạn

Đặt

A = \(x^4-4x^3-7x^2+16x-3\)

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

A \(=x^4+ax^3+x^2+bx^3+abx^2+bx-3x^2-3ax-3\)

A \(=x^4+\left(a+b\right)x^3+\left(1+ab-3\right)x^2+\left(b-3a\right)x-3\)

Đồng nhất 2 đa thức ta được :

\(\left\{{}\begin{matrix}a+b=-4\\-2+ab=-7\\b-3a=16\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=-5\\b=1\end{matrix}\right.\)

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

phân tích đa thức thành nhân tử

1) 4x^{8 + 1}

2) x^{7 +x5 +1}

3) x^{7 +x5 -1}

4) (x^{2-3x+5)2 - 7x(x2-3x+5)+12x2}

  bởi Đào Lê Hương Quỳnh 19/10/2018

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4) Ta có: \(\left(x^2-3x+5\right)^2-7x\left(x^2-3x+5\right)+12x^2\) (1)

Đặt \(x^2-3x+5=a\) => (1) trở thành:

\(a^2-7ax+12x^2\) = \(\left(a^2-3ax\right)-\left(4ax-12x^2\right)\)

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

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

Thay a = \(x^2-3x+5\) vào 2 ta được:

\(\left(x^2-6x+5\right)\left(x^2-7x+5\right)\)

= \(\left[\left(x^2-x\right)-\left(5x-5\right)\right]\left(x^2-7x+5\right)\)

= \(\left[x\left(x-1\right)-5\left(x-1\right)\right]\left(x^2-7x+5\right)\)

= \(\left(x-5\right)\left(x-1\right)\left(x^2-7x+5\right)\)

phân tích các đa thức sau thành nhân tử :

\(20x^7+7x-6\)

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

\(x^8+y^4+1\)

a)

\(x^3+x+2\)

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

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

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

b)

\(x^3-2x-1\)

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

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

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

c)

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

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

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

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

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

d)

\(x^3-3x^2y-9xy^2+5y^3\)

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

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

\(=\left(x-y\right)\left(x^2-4xy-5y^2\right)\)

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

\(x^4+2007x^2+2006x+2007\\ =x^4+x^3-x^3+x^2-x^2-x+2007x^2+2007x+2007\\ =\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(2007x^2+2007x+2007\right)\\ =x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2007\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^2-x+2007\right)\)

Phân tích đa thức thành nhân tử

a) x³ +x+2

b) x³-2x-1

c) x³+3x²-4

d) x³+3x²y-9xy²+5y³

b) (x²+8x-5)(x²+8x+1)-16

Đặt \(t=x^2+8x-5\) ta đc:

\(t\left(t+6\right)-16\)\(=t^2+6t-16\)

\(=t^2+8t-2t-16\)

\(=t\left(t+8\right)-2\left(t+8\right)\)

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

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

Phân tích thành nhân tử:

B= (a+b-2c)³ +(b+c-2a)³+(c+a-2b)3

Ta có

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

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

Áp dụng vào tổng ta có

\(\left(a+b-2c\right)^3+\left(b-c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt

\(M=\left(a+b-2c+b+c-2a+a+c-2b\right)^3\)

=> M=0

\(N=3\left(a+b-2c+b+c-2a\right)\left(b+c-2a+a+c-2b\right)\left(a+c-2b+a+b-2c\right)\)

\(\Rightarrow N=3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

Để ý : M+N=B

=> \(B=0+3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

=> \(B=3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

1,

20x⁷ + 7x - 6

= 20x⁷ - 8x + 15x - 6

= (20x⁷ - 8x )+ (15x - 6)

= 4x( 5x - 2 ) + 3( 5x - 2)

= ( 5x - 2 )( 4x + 3 )

2/

x⁴ - 5x²y² + 4y⁴

= x⁴ - 4x²y² - x²y² + 4y²

= (x⁴ - 4x²y² + 4y²) - x²y²

= ( x² - 2y²)² - x²y²

= ( x² - 2y²)² - (xy)²

= ( x² - 2y² - xy)( x² - 2y² + xy)

3/

x⁸ + y⁴ + 1

= ( x⁸ - x² ) + ( x⁴ + 1 + x² )

= x²( x⁶ - 1) + ( x⁴ + 1 + x² )

= x²( x³ - 1 )( x³ + 1) + ( x⁴ + 1 + x² )

= x² . ( x - 1 ).( x² + x + 1 ). ( x - 1).(x² - x + 1) + ( x⁴ + 1 + x² )

= x² . \(\left[\left(x-1\right)\left(x+1\right)\right].\)\(\left[\left(x^2+x+1\right)\left(x^2-x+1\right)\right]\)+ ( x⁴ + 1 + x² )

= x²(x² - 1 )\(\left[\left(x^2+1\right)^2-x^2\right]\)+ ( x⁴ + 1 + x² )

= x²(x² - 1 )( x⁴ + 2x² + 1 - x² ) + ( x⁴ + 1 + x² )

= x²(x² - 1 )( x⁴ +x² + 1 ) + ( x⁴ + 1 + x² )

= ( x⁴ - x² )( x⁴ +x² + 1 ) + ( x⁴ + 1 + x² )

= ( x⁴ +x² + 1 )( x⁴ - x² + 1 )

x⁴+2007x² + 2006x + 2007

Cách 2 : 

Đặt a + b - 2c = x ; b + c - 2a = y ; c + a - 2b = z 

=> x + y + z = a + b - 2c + b + c - 2a + c + a - 2b = 0

Do x + y + z = 0 => x + y = -z => (x+y)^3 = -z^3

Lại có : x^3 + y^3 + z^3 

= ( x + y )^3 - 3xy( x + y ) + z^3

= -z^3 - 3xy . ( - z ) + z^3

= 3xyz 

hay ( a + b - 2c )^3 + ( b + c - 2a)^3 + ( c + a - 2b)^3

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

Phân tích các đa thức thành nhân tử.

a) (x²  - 5x + 6 )(x² - 5x +2) - 5

b) (x² + 8x - 5 )(x² + 8x +1 ) -16

c) (x+2)(x+3)(x+4)(x+6) - 3/2.x²

^{Làm giúp mk nha.Đg cần gấp nek.........}

a ) (x^2 - 5x + 6)(x^2 - 5x + 2) - 5

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

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

= (x^2 - 5x + 4)^2 - 9

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

= (x^2 - 5x + 1)(x^2 - 5x + 7)

b ) (x^2 + 8x - 5)(x^2 + 8x + 1) - 16

= (x^2 + 8x - 2 - 3)(x^2 + 8x - 2 + 3) - 16

= (x^2 + 8x - 2)^2 - 9 - 16

= (x^2 + 8x - 2)^2 - 25

= (x^2 + 8x - 2 - 5)(x^2 + 8x - 2 + 5)

= (x^2 + 8x - 7)(x^2 + 8x + 3)