Phân tích đa thức thành nhân tử bằng phương pháp nhóm hạng tử:
a. 4x2 - 9y2 + 4x - 6y
b. x3 + y(1-3x2) + x(3y2-1) - y3
c. a2x + a2y - 7x - 7y
d. x(x+1)2 + x(x-5) - 5(x+1)2
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a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
a) x2 ( x+ 2y) -x -2y
= x2 ( x+ 2y) -(x+2y)
= (x2-1)(x+2y)
= (x-1)(x+1)(x+2y)
b)3x2- 3y2 -2 (x-y)2
= 3(x2-y2) -2 (x-y)2
= 3(x-y)(x+y)-2(x-y)(x-y)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)\\ =\left(x-y\right)\left(x+5y\right)\)
c) x2- 2x-4y2 - 4y
= (x2-4y2)-(2x+4y)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\left(x-2y-2\right)\)
d) x3 - 4x2 - 9x +36
= (x3+3x2)-(7x2+21x)+(12x+36)
= x2(x+3)-7x(x+3)+12(x+3)
=(x2-7x+12)(x+3)
\(=\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\left(x+3\right)\\ =\left[x\left(x-3\right)-4\left(x-3\right)\right]\left(x+3\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)
a) \(\left(x+2y\right)^2-\left(x-y\right)^2=\left(x+2y+x-y\right)\left(x+2y-x+y\right)\)
\(=\left(2x+y\right).3y\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)
\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)
c) \(9x^2-3x+2y-4y^2\)
\(=9x^2-4y^2-3x+2y\)
\(=\left(3x-2y\right)\left(3x+2y\right)-\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left[3x+2y-1\right]\)
d) \(4x^2-4xy+2x-y+y^2\)
\(=4x^2-4xy+y^2+2x-y\)
\(=\left(2x-y\right)^2+2x-y\)
\(=\left(2x-y\right)\left(2x-y+1\right)\)
e) \(x^3+3x^2+3x+1-y^3\)
\(=\left(x+1\right)^3-y^3\)
\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)
g) \(x^3-2x^2y+xy^2-4x\)
\(=x\left(x^2-2xy+y^2\right)-4x\)
\(=x\left(x-y\right)^2-4x\)
\(=x\left[\left(x-y\right)^2-4\right]\)
\(=x\left(x-y+2\right)\left(x-y-2\right)\)
a) (x + 2y)² - (x - y)²
= (x + 2y - x + y)(x + 2y + x - y)
= 3y(2x + y)
b) (x + 1)³ + (x - 1)³
= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]
= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)
= 2x(x² + 3)
c) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) x³ + 3x² + 3x + 1 - y³
= (x³ + 3x² + 3x + 1) - y³
= (x + 1)³ - y³
= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]
= (x - y + 1)(x² + 2x + 1 + xy + y + y²)
g) x³ - 2x²y + xy² - 4x
= x(x² - 2xy + y² - 4)
= x[(x² - 2xy + y²) - 4]
= x[(x - y)² - 2²]
= x(x - y - 2)(x - y + 2)
phân tích đa thức thành nhân tử bằng cách nhóm hạng tử
1) x2 - y2 - 2x - 2y
2) 3x2 - 3y2 - 2(x - y)2
1) \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x-y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
2) \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
1) x² - y² - 2x - 2y
= (x² - y²) - (2x + 2y)
= (x - y)(x + y) - 2(x + y)
= (x + y)(x - y - 2)
2) 3x² - 3y² - 2(x - y)²
= (3x² - 3y²) - 2(x - y)²
= 3(x² - y²) - 2(x - y)²
= 3(x - y)(x + y) - 2(x - y)²
= (x - y)[3(x + y) - 2(x - y)]
= (x - y)(3x + 3y - 2x + 2y)
= (x - y)(x + 5y)
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
c) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)
d) \(y^2\left(x-1\right)-7y^3+7xy^3\)
\(=y^2\left(x-1-7y+7xy\right)\)
\(=y^2\left[\left(x-1\right)-7y\left(1-x\right)\right]=y^2\left(x-1\right)\left(1+7y\right)\)
a)
\(xy+y^2-x-y\\ =\left(xy-x\right)+\left(y^2-y\right)\\ =x\left(y-1\right)+y\left(y-1\right)\\ =\left(y-1\right)\left(x+y\right)\)
Bài 1 yêu cầu gì em?
Bài 2:
\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)
Bài 1:
a) \(3xy+6y\)
\(=3y\left(x+2\right)\)
b) \(3x^2+9x\)
\(=3x\left(x+3\right)\)
c) \(6x-9y^2\)
\(=3\left(2x-3y^2\right)\)
d) \(10xy^2-6x^2y\)
\(=2xy\left(5y-3x\right)\)
Bài 2:
a) \(x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x+5\right)\)
b) \(3x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(x+1\right)\)
\(=3\left(x+1\right)^2\)
c) \(x\left(x-3\right)+xy\left(x-3\right)\)
\(=\left(x+xy\right)\left(x-3\right)\)
\(=x\left(1+y\right)\left(x-3\right)\)
d) \(2x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(2x-6\right)\left(x-2\right)\)
\(=2\left(x-3\right)\left(x-2\right)\)
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.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
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: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
Lời giải:
a) $4x^2-9y^2+4x-6y=(4x^2+4x)-(9y^2+6y)$
$=(4x^2+4x+1)-(9y^2+6y+1)$
$=(2x+1)^2-(3y+1)^2=(2x+1-3y-1)(2x+1+3y+1)=(2x-3y)(2x+3y+2)$
b)
$x^3+y(1-3x^2)+x(3y^2-1)-y^3=x^3+y-3x^2y+3xy^2-x-y^3$
$=(x^3-3x^2y+3xy^2-y^3)-(x-y)=(x-y)^3-(x-y)=(x-y)[(x-y)^2-1]$
$=(x-y)(x-y-1)(x-y+1)$
c)
$a^2x+a^2y-7x-7y=a^2(x+y)-7(x+y)=(a^2-7)(x+y)$
d)
$x(x+1)^2+x(x-5)-5(x+1)^2=x(x+1)^2-5(x+1)^2+x(x-5)$
$=(x+1)^2(x-5)+x(x-5)=(x-5)[(x+1)^2+x]=(x-5)(x^2+3x+1)$