a/ x² + 6x + 9 = (x + 3)² = (x + 3)(x + 3)

b/ 10x - 25 - x² = -x² + 10x - 25 = -(x² -10x + 25) = -(x - 5)² = -(x - 5)(x - 5)

c/ \(8x^3+\frac{1}{8}=\left(2x\right)^3+\left(\frac{1}{2}\right)^3=\left(2x+\frac{1}{2}\right)\left(4x^2-x+\frac{1}{4}\right)\)

\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

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= (5x-25) + (5x - x²)

= 5(x-5) + x(5-x)

= 5(x-5) - x(x-5)

= (5 - x)(x - 5)

\(10x-25-x^2=-\left(x^2-2\cdot x\cdot5+5^2\right)=-\left(x-5\right)^2\)

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

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

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

                                      = 4x(2x+1)

a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

c) \(=\left(4x-3y\right)\left(4x+3y\right)\)

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

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

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

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

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

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

\(a,=5\left(x-2y\right)\\ b,=3xy\left(x-2y\right)\\ c,=\left(x-y\right)\left(x+3\right)\\ d,=\left(x-1\right)\left(2x-4x^2\right)=2x\left(1-2x\right)\left(x-1\right)\\ e,=\left(x-2y\right)^2\\ f,=\left(3x-4y\right)\left(3x+4y\right)\\ g,=\left(x-3\right)\left(x^2+3x+9\right)\)

a. 5x - 10y 

= 5(x - 2y)

b. 3x²y - 6xy²

= 3xy(x - 2y)

c. x(x - y) - 3(y - x)

= x(x - y) + 3(x - y)

= (x + 3)(x - y)

d. 2x(x - 1) + 4x²(1 - x)

= 2x(x - 1) - 4x²(x - 1)

= (2x - 4x²)(x - 1)

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

e. x² - 4xy + 4y²

= (x - 2y)²

f. 9x² - 16y²

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

g. x³ - 27

= (x - 3)(x² + 3x + 9)

Câu 1:8-x^3=2^3-x^3=(2-x)(4+2x+x^2)

Câu 2:Ta có:x^2-5x+4

=(x^2-2x5/2+25/4)-9/4

=(x-5/2)^2-(3/2)^2

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

=(x-4)(x-1)

->đa thức B là:(x-4)

->hệ số tự do của đa thức B là:-4

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

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

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

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

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