phân tích:
a) x^4 - 4x^2 + 8x - 8
b) a^2 + b^2 + a^2b^2 + ab - a - b
c) x^2y + xy^2 + x^2z + y^2z + yz^2 + 2xyz
Phân tích các đa thức thành nhân tử
a)x^3-4x^2+8x-8
b)a^2+b^2-a^2b^2+ab-a-b
c)x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz
a. bc(b+c)+ca(c-a)-ab(a+b)
b. 2a^2b+4ab^2-a^2c+a^2-4b^2c+2bc^2-4abc
c. y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2
d. x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz
phân tích đa thức thành nhân tử
1)bc(b+c)+ca(c-a)-ab(a+b)
2)\(2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc\)
3)y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2
4)\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
Phân tích các biểu thức sau thành nhân tử:
1) A=\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
2) B=\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
3) C=\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
4) D=\(2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4a^2c\)
5) \(E=y\left(x-2z\right)^2+8xyz+x\left(y-2z\right)^2-2z\left(x+y\right)^2\)
6)F=\(8x^3\left(y+z\right)-y^3\left(z+2x\right)-z^3\left(2x-y\right)\)
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\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
Phân tích đa thức thành nhân tử
a) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
Hóng cao nhân , CTV vô đê , tận 30 người cơ mà
a) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=\left(x+y+z\right)\left(xy+xz\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2+xyz\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z+x\right)\)
\(=\left(x+y+z\right)\left(xy+xz+yz\right)\)
P/s: Sai sót xin bỏ qua.
phân tích thành nhân tử :
a) \(a^4+a^3+a^{3b}+a^{2b}\)
b) \(a^3+4a^2+4a+3\)
c)\(a^3+3a^2+4a+12\)
d) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(a^4+a^3+a^{3b}+a^{2b}\)
\(=a\left(a^3+a^2+1^{3b}+1^{2b}\right)\)
\(a^3+3a^2+4a+12\)
\(=a^2\left(a+3\right)+4\left(a+3\right)\)
\(=\left(a^2+4\right)\left(a+3\right)\)
phân tích đa thức thành nhân tử:
a. \(3xyz+x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)\)
b.\(x^8y^8+x^4y^4+1\)
c.\(x^2y+xy^2+xz^2+x^2z+y^2z+yz^2+2xyz\)
b \(x^8y^8+x^4y^4+1=x^8y^8+2x^4y^4+1-x^4y^4=\left(x^4y^4\right)^2+2x^4y^4+1-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2=\left(x^4y^4-x^2y^2+1\right)\left(x^4y^4+x^2y^2+1\right)\)
c \(x^2y+xy^2+xz^2+x^2z+y^2z+yz^2+2xyz=\left(x^2y+x^2z+xyz+xy^2\right)+\left(xz^2+yz^2+xyz+y^2z\right)\)
\(=x\left(xy+xz+yz+y^2\right)+z\left(xz+yz+xy+y^2\right)=\left(x+z\right)\left(xy+xz+yz+y^2\right)\)
\(=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)
a \(3xyz+x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)=3xyz+xy^2+xz^2+x^2y+yz^2+x^2z+y^2z\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xy^2+xyz+y^2z\right)+\left(xyz+xz^2+yz^2\right)\)
\(=x\left(xy+xz+yz\right)+y\left(xy+xz+yz\right)+z\left(xy+xz+yz\right)=\left(x+y+z\right)\left(xy+xz+yz\right)\)
Rút gọn các biểu thức sau:
a) A= 1/3xy + 4xy - 2xy
b) B=-xy^2 + 3/2xy^2 + 4/3xy^2
c) C= (2xy)^2 + 2/3x^2y^2 - 4/3xyx
d) D= x. (3xy^2z) + 4x^2y^2z - 8x^2y . yz
a: =xy(1/3+4-2)=7/3xy
b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2
c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2
d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z
phân tích các đa thức sau thành nhân tử
a) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz.\)
\(=x^2.\left(y+z\right)+yz.\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)
\(=\left(y+z\right).\left(x^2+yz\right)+x\left(y^{^2}+z^2+2yz\right)\)
\(=\left(y+z\right).\left[x.\left(x+2\right)+y.\left(x+2\right)\right]\)
\(=\left(y+z\right).\left(x+z\right).\left(x+y\right)\)