Phân tích đa thức sau thành nhân tử: x3 + 2x2y + xy2 – 9x
phân tích đa thức sau thành nhân tử: x3+2x2y+xy2-9x
\(=x\left(x^2+2xy+y^2-9\right)\)
=x(x+y-3)(x+y+3)
\(x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-3^2\right]=x\left(x+y-3\right)\left(x+y+3\right)\)
Phân tích đa thức thành nhân tử: x 3 – 9 x + 2 x 2 y + x y 2
A. x. (x - y + 3).(x + y - 3)
B. x. (x + y + 3).(x + y - 3)
C. x. (x - y + 3).(x - y - 1)
D. x. (x + y + 1).(x - y - 3)
Phân tích đa thức sau thành nhân tử:
a) x3 + 2x2y + xy2 - 9x b) 5x2 - 10xy + 5y2
a) x3 + 2x2y + xy2 - 9x
= x(x2 + 2xy + y2 - 9)
= x(x+y)2 - 9
= x(x + y - 3) ( x + y + 3).
b) 5x2 - 10xy + 5y2
= 5(x2 - 2xy + y2)
= 5(x-y)2
Có sai thì xin lỗi ạ
Phân tích đa thức sau thành nhân tử:
a ) x 3 + 2 x 2 y + x y 2 – 4 x
a) x3 + 2x2y + xy2 – 4x = x(x2 + 2xy + y2– 4) = x[(x+y)2-4]
= x(x + y + 2)(x + y – 2)
Phân tích các đa thức sau thành nhân tử:
a/ 2x3 + 3x2 + 2x +3 b/ x2 – x – 12 c/ 4x2 –( x2 + 1)2
d/ 4xy2 – 12x2y + 8xy e/ x2 + x – 6 f/ x3 + 2x2y + xy2 – 4xz2
g/ x3 – 2x2y + xy2 – 25x h/ x2 – 2x – 3 i/ x3 – 3x2 – 9x + 27
a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+1\right)\)
b: \(=\left(x-4\right)\left(x+3\right)\)
e: =(x+3)(x-2)
a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)
b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)
c) \(=\left(2x\right)^2-\left(x^2+1\right)^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=4xy\left(y-3x+2\right)\)
e) \(=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)
f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)
g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)
h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)
i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)
2x2y+xy2-xy giúp mik ạ phân tích các đa thức sau thành phân tử bằng cách đặt nhân tử chung
=xy*2x+xy*y-xy*1
=xy(2x+y-1)
phân tích các đa thức sao thành nhân tử
a.x2-10x+25
b.(2x-3)2-x2
c.x3+2x2y+xy2-9x
a) x² - 10x + 25
= x² - 2.x.5 + 5²
= (x - 5)²
b) (2x - 3)² - x²
= (2x - 3 - x)(2x - 3 + x)
= (x - 3)(3x - 3)
= 3(x - 3)(x - 1)
c) x³ + 2x²y+ xy² - 9x
= x(x² + 2xy + y² - 9)
= x[(x + y)² - 3²]
= x(x + y - 3)(x + y + 3)
a/\(x^2-10x+25\)
\(=\left(x-5\right)^2\)
b/\(\left(2x-3\right)^2-x^2\)
\(=\left(2x-3-x\right)\left(2x-3+x\right)\)
\(=\left(x-3\right)\left(3x-3\right)\)
\(=3\left(x-3\right)\left(x-1\right)\)
c/\(x^3+2x^2+xy^2-9x\)
\(=x\left(x^2+2x+y^2-9\right)\)
\(=x\left[\left(x+y\right)^2-3^2\right]\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
#sdboy2mai
Phân tích đa thức thành nhân tử:
+)5x2y2+15x2+30xy2
+)(x-2)(x-3)+4-x2
+)x2-7x+12
+)x3-2x2y+xy2-9x
+)x2-25+y2+2xy
+)x2-x-12
+)5x25xy-x-y
+)12y(2x-5)+6xy(5-2x)
+)16x2+24x-8xy-6y+y2
+)(x+3)(x+6)(x+9)(x+12)+81
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
Bài 1: Phân tích đa thức thành nhân tử
a) a/ x2 – 2x
b) 2bx – 3ay – 6by + ax
c) x3 +2x2y + xy2 – 4x
d) 4 - x2 – 2xy – y2
đ) 5x2 + 3(x + y)2 – 5y2
e/ 6x2y – 9x
b/ 4x3 – 4x2y + xy2 – 16 x
f) x2 + (2x +y)y – z2
\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)
a,=x(x−2)b,=2b(x−3y)+a(x−3y)=(a+2b)(x−3y)c,=x(x2+2xy+y2−4)=x[(x+y)2−4]=x(x+y+2)(x+y−2)d,=4−(x+y)2=(2−x−y)(2+x+y)đ,=5(x−y)(x+y)+3(x+y)2=(x+y)(5x−5y+3x+3y)=(x+y)(8x−2y)=2(4x−y)(x+y)e,=3x(2xy−3)b,=x(4x2−4xy+y2−4)=x[(2x−y)2−4]=x(2x−y−2)(2x−y+2)f,=(x+y)2−z2=(x+y−z)(x+y+z)