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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: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(4x^2+4x+1=\left(2x+1\right)^2\)
g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c: \(2x-1-x^2\)
\(=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\)
d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)
g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)
\(=\left(5-x\right)\left(5+3x\right)\)
h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)
\(=\left(x+1+2x-1\right)\left(x+1-2x+1\right)\)
\(=3x\left(-x+2\right)\)
i: \(=x^2y^2-4xy+4-3\)
\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)
k: \(=y^2-\left(x-1\right)^2\)
\(=\left(y-x+1\right)\left(y+x-1\right)\)
l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)
m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)
Bài 1:
a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)
c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
Bài 1 : Phân tích các đa thức sau thành nhân tử :
a) 8x3 - 64
=(2x)3 + 43
=(2x+4)(4x2 - 8x + 16)
c) 125x3 + 1
=5x3 + 13
=(5x+1)(25x2 +5x+1)
d) 8x3 - 27
=(2x)3 - 33
=(2x - 3)(2x2 + 6x + 9)
e) 1 + 8x6y3
=1 + (2x2y)3
=(1 + 2x2y)(4x4y2 -2x2y + 1)
f) 125x3 + 27y3
=(5x)3 + (3y3)
=(5x + 3y)(25x2 - 15xy + 9y2)
Bài 1
a) \(8x^3-64\)
\(=\left(2x\right)^3-4^3\)
\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)
c) \(125x^3+1\)
\(=\left(5x\right)^3+1^3\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right)\)
d) \(8x^3-27\)
\(=\left(2x\right)^3-3^3\)
\(=\left(2x-3\right)\left(4x^2+6x+9\right)\)
e) \(1+8x^6x^3\)
\(=1^3+\left(2x^2y\right)^3\)
\(=\left(1+2x^2y\right)\left(1-2x^2y+4x^4y^2\right)\)
f) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)