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a) \(x^4+8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)
Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)
\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)
\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)
\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=3\left(x^2-10x+25\right)=3\left(x-5\right)^2\)
b) \(=x\left(x+y\right)+8\left(x+y\right)=\left(x+y\right)\left(x+8\right)\)
c) \(=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
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a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)
b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)
d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)
e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2x+1-16=\left(x+1\right)^2-4^2=\left(x+1-4\right).\left(x+1+4\right)=\left(x-3\right).\left(x+5\right)\)
\(x^2+2x+1-16=\left(x^2+2x+1\right)-4^2=\left(x+1\right)^2-4^2=\left(x+1-4\right)\left(x+1+4\right)=\left(x-3\right)\left(x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
2xy – x2 – y2 + 16 (Có 2xy ; x2 ; y2, ta liên tưởng đến HĐT (1) hoặc (2))
= 16 – (x2 – 2xy + y2)
= 42 – (x – y)2 (xuất hiện hằng đẳng thức (3))
= [4 – (x – y)][4 + (x - y)]
= (4 – x + y)(4 + x – y).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x+4\right)^2\)
\(x^2+8x+16=\left(x+4\right)^2\)