Phân tích các đa thức sau thành nhân tử:
a) xy – 3x + 2y – 6
b) x2y + 4xy + 4y – y3
c) x2 + y2 + xz + yz + 2xy
d) x3 + 3x2 – 3x – 1
Phân tích các đa thức sau thành nhân tử:
a/ x( 3- x) – x + 3 b/ 3x2 – 5x – 3xy + 5y c/ x2 – xy – 10x + 10y
d/ 2xy+ x2 + y2 - 16 e/ x2 – y2 – 4x – 4y f/ 9 – 4x2 + 4xy – y2
g/ y3 – 2xy2 + x2y h/ x3 – 3x2 – 4x + 12 i/ x( x- y) + x2 – y2
a: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
Phân tích các đa thức sau thành nhân tử:
a,x3+4x-5
b,x3-3x2+4
c,x3+2x2+3x+2
d,x2+2xy+y2+2x-2y-3
e,(x2+3x)2-2(x2+3x)-8
f,(x2+4x+10)2-7(x2+4x+11)+7
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)\left(x^2+3x-4\right)+2\left(x^2+3x-4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x+4\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)
Phân tích các đa thức sau thành nhân tử:
a/ x2 – 3x – 4xy + 12y b/ x3 – 4x2 + 4x -1
c/ x – y – ax + ay d/ x2 – 4 + ( x + 2)2
e/x3 + x2y – x2z – xyz f/ x2 – y2 – 2x – 2y
a: \(=x\left(x-3\right)-4y\left(x-3\right)\)
=(x-3)(x-4y)
d: \(=\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2+x+2\right)\)
=2x(x+2)
\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)
Phân tích các đa thức sau thành nhân tử:
a) 3x2 + xy - 4y2
b) x8 - 5x4 + 4
c) x3 + 3x2 + 3x - 7
Lời giải:
a.
$3x^2+xy-4y^2=(3x^2-3xy)+(4xy-4y^2)=3x(x-y)+4y(x-y)=(x-y)(3x+4y)$
b.
$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$
$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$
$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$
$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$
c.
$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$
$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$
$=(x-1)(x^2+4x+7)$
a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)
\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)
b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)
\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)
\(=(x-2)(x+2)(x^2-1)(x^2+1)=(x-2)(x+2)(x-1)(x+1)(x^2+1)\)
c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)
\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)
a: \(3x^2+xy-4y^2\)
\(=3x^2+4xy-3xy-4y^2\)
\(=x\left(3x+4y\right)-y\left(3x+4y\right)\)
\(=\left(3x+4y\right)\left(x-y\right)\)
b: \(x^8-5x^4+4\)
\(=x^8-x^4-4x^4+4\)
\(=x^4\left(x^4-1\right)-4\left(x^4-1\right)\)
\(=\left(x^4-4\right)\left(x^4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^2-2\right)\left(x^2+2\right)\)
phân tích các đa thức sau thành nhân tử: a) 4x(2x - 3y) - 8y(3y - 2x) b) 4x2 - 4xy + y2 - 9z2 c) x2y + yz + xy2 + xz d) (1 - x2)x2 - 16x2 - 16
Bạn thử xem lại đề câu d nhé.
a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)
\(=4x\left(2x-3y\right)+8y\left(2x-3y\right)\)
\(=4\left(2x-3y\right)\left(x+2y\right)\)
b) Ta có: \(4x^2-4xy+y^2-9z^2\)
\(=\left(2x+y\right)^2-\left(3z\right)^2\)
\(=\left(2x+y+3z\right)\left(2x+y-3z\right)\)
c) Ta có: \(x^2y+yz+xy^2+xz\)
\(=xy\left(x+y\right)+z\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+z\right)\)
Phân tích đa thức thành nhân tử:
a) x3 - 2x2 - 2x - 4
b) xy + 1 - x - y
c) x2 - 4xy + 4y2 - 4y
d) 16 - x2 + 2xy - y2
\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
Phân tích đa thức thành nhân tử:
a.10x2y – 20xy2 b. x2 – y2 + 10y – 25 c. x2 – y2 + 3x – 3y
d. x3 + 3x2 – 16x – 48 e. 9x3 + 6x2 + x f. x4 + 5x3 + 15x – 9
\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)
\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)
Phân tích các đa thức sau thành nhân tử:
a) 3x4y - 12x2y3
b) x2 - y2 - 8y -16
c) x3 +3x2 + 4x +12
d) 3x2 - 6xy + 3y2 - 27
\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)
\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)
\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)
\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)
Phân tích các đa thức sau thành nhân tử:
a) 6x3y-9x2y2+3xy
b) x2-3x+xy-3y
c) x2-y2-4x+4
b) x2-3x+xy-3y
=\(\left(x^2+xy\right)-\left(3x+3y\right)\)
=\(x\left(x+y\right)-3\left(x+y\right)\)
=\(\left(x-3\right)\left(x+y\right)\)
c) x2-y2-4x+4
=(\(x^2-4x+4\))\(-y^2\)
=\(\left(x-2\right)^2\) \(-y^2\)
=(\(x-y-2\)) \(\left(x+y-2\right)\)
a)6x3y-9x2y2+3xy
=\(3xy\left(2x^2-3xy+1\right)\)