Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

a: \(x^2-y^2+3x+3y\)

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

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

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

b: Sửa đề: \(x^2-4y^2+4x+4\)

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

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

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

1: \(12x^2y-18xy^2-30y^2\)

\(=6y\cdot2x-6y\cdot3xy-6y\cdot5y\)

\(=6y\left(2x-3xy-5y\right)\)

2: \(25-\left(3-x\right)^2\)

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

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

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

3: \(9-x^2+2xy-y^2\)

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

\(=9-\left(x-y\right)^2\)

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

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

4: \(xy-xz+y-z\)

\(=\left(xy-xz\right)+\left(y-z\right)\)

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

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

5: \(x^3-16x\)

\(=x\cdot x^2-x\cdot16\)

\(=x\left(x^2-16\right)\)

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

6: \(81-x^2+4xy-4y^2\)

\(=81-\left(x^2-4xy+4y^2\right)\)

\(=81-\left(x-2y\right)^2\)

\(=\left[9-\left(x-2y\right)\right]\left[9+\left(x-2y\right)\right]\)

\(=\left(9-x+2y\right)\left(9+x-2y\right)\)

7 \(32x^3-16x^2-19xy^2+2x\)

\(=2x\cdot16x^2-2x\cdot8x-2x\cdot9,5y^2+2x\cdot1\)

\(=2x\left(16x^2-8x+1-9,5y^2\right)\)

\(=2x\left[\left(16x^2-8x+1\right)-\left(y\cdot\sqrt{9,5}\right)^2\right]\)

\(=2x\left[\left(4x-1\right)^2-\left(\dfrac{y\sqrt{38}}{2}\right)^2\right]\)

\(=2x\left(4x-1-\dfrac{y\sqrt{38}}{2}\right)\left(4x-1+\dfrac{y\sqrt{38}}{2}\right)\)

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

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

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

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

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

9: \(16x^3+54y^3\)

\(=2\cdot8x^3+2\cdot27y^3\)

\(=2\left(8x^3+27y^3\right)\)

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

\(=2\left(2x+3y\right)\left[\left(2x\right)^2-2x\cdot3y+\left(3y\right)^2\right]\)

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

10: \(x^2-5x+6\)

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

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

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

Sửa đề: \(16\left(x-1\right)^2-9\left(2x+1\right)^2=0\)

=>\(\left[4\left(x-1\right)\right]^2-\left[3\left(2x+1\right)\right]^2=0\)

=>\(\left(4x-4\right)^2-\left(6x+3\right)^2=0\)

=>\(\left(4x-4-6x-3\right)\left(4x-4+6x+3\right)=0\)

=>\(\left(-2x-7\right)\left(10x-1\right)=0\)

=>\(\left(2x+7\right)\left(10x-1\right)=0\)

=>\(\left[{}\begin{matrix}2x+7=0\\10x-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=-7\\10x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{1}{10}\end{matrix}\right.\)

a: \(x^2-4x\)

\(=x\cdot x-x\cdot4\)

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

b: \(2x^3-2x\)

\(=2x\cdot x^2-2x\cdot1\)

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

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

a: Ta thấy khi x=1 thì \(x^3-3x+2=1^3-3\cdot1+2=1-3+2=0\)

=>x=1 là nghiệm của đa thức \(x^3-3x+2\)

Do đó, ta sẽ chia x^3-3x+2 cho x-1

Từ phép chia trên, ta sẽ có: \(x^3-3x+2=\left(x-1\right)\left(x^2-x-2\right)\)

mà ta lại có: \(x^2-x-2=x^2-2x+x-2=\left(x-2\right)\left(x+1\right)\)

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

b: Khi x=1 thì \(9x^3+3x^2-8x-4=9+3-8-4=0\)

=>x=1 là nghiệm của đa thức \(9x^3+3x^2-8x-4\)

Do đó, ta sẽ chia đa thức \(9x^3+3x^2-8x-4\) cho x-1

Từ phép chia trên, ta sẽ có: \(9x^3+3x^2-8x-4=\left(x-1\right)\left(9x^2+12x+4\right)\)

\(=\left(x-1\right)\left[\left(3x\right)^2+2\cdot3x\cdot2+2^2\right]\)

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

c: Khi \(x^2=\dfrac{1}{3}\) thì \(6x^4-11x^2+3=6\left(x^2\right)^2-11x^2+3=6\cdot\left(\dfrac{1}{3}\right)^2-11\cdot\dfrac{1}{3}+3=6\cdot\dfrac{1}{9}+3-\dfrac{11}{3}=\dfrac{2}{3}+3-\dfrac{11}{3}=0\)

=>\(x^2=\dfrac{1}{3}\) là nghiệm của đa thức \(6x^4-11x^2+3\)

=>Ta sẽ chia đa thức \(6x^4-11x^2+3\) cho \(x^2-\dfrac{1}{3}\)

Từ phép chia trên, ta sẽ có:

\(6x^4-11x^2+3=\left(x^2-\dfrac{1}{3}\right)\left(6x^2-9\right)\)

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

a: \(x^4+324\)

\(=x^4+36x^2+324-36x^2\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot18+18^2-36x^2\)

\(=\left(x^2+18\right)^2-36x^2\)

\(=\left(x^2-6x+18\right)\left(x^2+6x+18\right)\)

b: \(4y^4+1\)

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

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

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

c: \(x^4y^4+4\)

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

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

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

\(a,x^4+324\\=[(x^2)^2+2\cdot x^2\cdot18+18^2]-2\cdot x^2\cdot18\\=(x^2+18)^2-36x^2\\=(x^2+18)^2-(6x)^2\\=(x^2+18-6x)(x^2+18+6x)\\b,4y^4+1\\=[(2y^2)^2+2\cdot2y^2\cdot1+1^2]-4y^2\\=(2y^2+1)^2-(2y)^2\\=(2y^2+1-2y)(2y^2+1+2y)\)

\(c,x^4y^4+4\\=[(x^2y^2)^2+2\cdot x^2y^2\cdot2+2^2]-4x^2y^2\\=(x^2y^2+2)^2-(2xy)^2\\=(x^2y^2+2-2xy)(x^2y^2+2+2xy)\\Toru\)

a: \(x^3-3x+2\)

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

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

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

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

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

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

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

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

b: \(9x^3+3x^2-8x-4\)

\(=9x^3-9x^2+12x^2-12x+4x-4\)

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

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

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

c: \(6x^4-11x^2+3\)

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

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

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

\(a,x^3-3x+2\\=x^3-x-2x+2\\=x(x^2-1)-2(x-1)\\=x(x-1)(x+1)-2(x-1)\\=(x-1)[x(x+1)-2]\\=(x-1)(x^2+x-2)\\=(x-1)(x^2-x+2x-2)\\=(x-1)[x(x-1)+2(x-1)]\\=(x-1)^2(x+2)\)

\(b,9x^3+3x^2-8x-4\\=9x^3-9x^2+12x^2-12x+4x-4\\=9x^2(x-1)+12x(x-1)+4(x-1)\\=(x-1)(9x^2+12x+4)\\=(x-1)[(3x)^2+2\cdot 3x\cdot2+2^2]\\=(x-1)(3x+2)^2\)

\(c,6x^4-11x^2+3\\=6x^4-2x^2-9x^2+3\\=2x^2(3x^2-1)-3(3x^2-1)\\=(3x^2-1)(2x^2-3)\\\text{#}Toru\)

`#3107.101107`

`7.`

`a)`

`x^3 - 3x + 2`

`= x^3 - 2x - x + 2`

`= (x^3 - x) - (2x - 2)`

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

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

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

`= (x^2 + x - 2)(x - 1)`

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

`= (x - 1)^2 (x + 2)`

`b)`

`9x^3 + 3x^2 - 8x - 4`

`= 9x^3 + 12x^2 - 9x^2 + 4x - 12x - 4`

`= (9x^3 + 12x^2 + 4x) - (9x^2 + 12x + 4)`

`= x(9x^2 + 12x + 4) - (9x^2 + 12x + 4)`

`= (x - 1)(9x^2 + 12x + 4)`

`= (x - 1)(3x + 2)^2`

`c)`

`6x^4 - 11x^2 + 3`

`= 6x^4 - 2x^2 - 9x^2 + 3`

`= (6x^4 - 2x^2) - (9x^2 - 3)`

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

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

a: \(2x^2+5xy-3y^2\)

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

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

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

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

b: \(2a^2+3ab-5b^2\)

\(=2a^2-2ab+5ab-5b^2\)

\(=\left(2a^2-2ab\right)+\left(5ab-5b^2\right)\)

\(=2a\left(a-b\right)+5b\left(a-b\right)\)

\(=\left(a-b\right)\left(2a+5b\right)\)

c: \(x^3+9x^2y+20xy^2\)

\(=x\cdot x^2+x\cdot9xy+x\cdot20y^2\)

\(=x\left(x^2+9xy+20y^2\right)\)

\(=x\left(x^2+4xy+5xy+20y^2\right)\)

\(=x\left[\left(x^2+4xy\right)+\left(5xy+20y^2\right)\right]\)

\(=x\left[x\left(x+4y\right)+5y\left(x+4y\right)\right]\)

\(=x\left(x+4y\right)\left(x+5y\right)\)

a) \(x^2+5x-6\)

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

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

\(=x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(x-1\right)\left(x-6\right)\)

b) \(a^2+4a+3\)

\(=a^2+a+3a+3\)

\(=\left(a^2+a\right)+\left(3a+3\right)\)

\(=a\left(a+1\right)+3\left(a+1\right)\)

\(=\left(a+1\right)\left(a+3\right)\)

c) \(16y-5y^2-3\)

\(=-5y^2+y+15y-3\)

\(=\left(-5y^2+y\right)+\left(15y-3\right)\)

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

\(=\left(5y-1\right)\left(3-y\right)\)

a: \(x^2+5x-6\)

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

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

\(=\left(x+6\right)\left(x-1\right)\)

b: \(a^2+4a+3\)

\(=a^2+a+3a+3\)

\(=a\left(a+1\right)+3\left(a+1\right)\)

\(=\left(a+1\right)\left(a+3\right)\)

c: \(16y-5y^2-3\)

\(=-5y^2+15y+y-3\)

\(=-5y\cdot\left(y-3\right)+\left(y-3\right)\)

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

1: Đa thức này ko phân tích được nha bạn

2: \(x^2+8x+7\)

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

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

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

3: \(x^2-6x-16\)

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

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

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

4: \(4x^2-8x+3\)

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

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

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

5: \(3x^2-11x+6\)

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

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

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