Phân tích đa thức thành nhan tử
a) x2y + xy2 + xz + xz2 + y2z + yz2 + 2xyz
b) x16 + x8 - 2
c) ( a + 1)(a + 2)( a +3)(a + 4) + 1
K
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T
26 tháng 11 2023
a) \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(5a+6b\right)\left(7-2b\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)
\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)
\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)
\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)
\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 11 2023
a, 70a + 84b - 20ab - 24b2
= 14.(5a + 6b) - 4b(5a + 6b)
= (5a + 6b).(14 - 4b)
26 tháng 11 2023
a: \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(7-2b\right)\left(5a+6b\right)\)
b: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2\right)+\left(y^2z+yz^2\right)+3xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+3xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+2xyz+xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2+2yz\right)+yz\left(y+z+x\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)^2+yz\left(y+z+x\right)\)
\(=\left(y+z\right)\cdot x\left(x+y+z\right)+yz\left(y+z+x\right)\)
\(=\left(y+z+x\right)\cdot\left(xy+xz+yz\right)\)
c: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2+2xyz\right)+\left(y^2z+yz^2\right)\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2+2xz\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y+z\right)^2\)
\(=\left(y+z\right)\left(x^2+yz+xy+xz\right)\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
LM
1
H9
HT.Phong (9A5)
CTVHS
14 tháng 9 2023
3) \(x^2\left(x+2y\right)-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x^2-1\right)\left(x+2y\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2y\right)\)
4) \(x^3-4x^2-9x+36\)
\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)
\(=x^2\cdot\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x+3\right)\left(x-3\right)\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
15 tháng 9 2023
\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
H
0
N
2
26 tháng 10 2021
\(a,=\left(x-y\right)\left(x+y\right)+11\left(x-y\right)=\left(x-y\right)\left(x+y+11\right)\\ b,=\left(x+z\right)\left(x^2-xz+z^2\right)+y\left(x^2+z^2-xz\right)\\ =\left(x^2-xz+z^2\right)\left(x+y+z\right)\)
a, \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+xy^2\right)+\left(x^2z+xyz\right)+\left(xz^2+yz^2\right)+\left(yz^2+xyz\right)\)
\(=xy\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+yz\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz+yz+z^2\right)=\left(x+y\right)[x\left(y+z\right)+z\left(y+z\right)]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
b,\(x^{16}+x^8-2=x^{16}+x^8-1-1=\left(x^{16}-1\right)+\left(x^8-1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)+\left(x^8-1\right)=\left(x^8-1\right)\left(x^8+2\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+2\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)c,\(A=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)+1\)\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
Đặt \(a^2+5a+5=k\) thế vào biểu thức A ta có:
\(A=\left(k-1\right)\left(k+1\right)+1=k^2-1+1=k^2=\left(a^2+5a+5\right)^2\)
C làm đi mày oi