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Phân Tích Đa Thức Thành Nhân Tử:
A=(x2+x)2+5y(x2+x)+6y2
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NT
1
NT
1
7 tháng 9 2021
a: \(16x^5-25x^3\)
\(=x^3\left(16x^2-25\right)\)
\(=x^3\left(4x-5\right)\left(4x+5\right)\)
b: \(x^2-2xy+y^2-16\)
\(=\left(x-y\right)^2-16\)
\(=\left(x-y-4\right)\left(x-y+4\right)\)
c: \(x^2-5y-xy+5x\)
\(=x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+5\right)\)
d: \(x^2\left(x^2+4\right)-x^2+4\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
26 tháng 10 2021
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: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
26 tháng 10 2021
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) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
2
24 tháng 4 2022
`a) 8x^2 - 8xy - 4x + 4y`
`= 8x ( x - y ) - 4 ( x - y )`
`= ( x - y ) ( 8x - 4 )`
__________________________
`b) x^3 + 10x^2 + 25x - xy^2`
`=x ( x^2 + 10x + 25 ) - xy^2`
`= x ( x + 5 )^2 - xy^2`
`= x [ ( x + 5 )^2 - y^2 ]`
`= x ( x + 5 - y ) ( x + 5 + y )`
________________________________
`c) x^2 + x - 6`
`= x^2 + 3x - 2x - 6`
`= x ( x + 3 ) - 2 ( x + 3 )`
`= ( x + 3 ) ( x - 2 )`
_______________________________
`d) 2x^2 + 4x - 16`
`= 2x^2 - 4x + 8x - 16`
`= 2x ( x - 2 ) + 8 ( x - 2 )`
`= ( x - 2 ) ( 2x + 8 )`
LS
24 tháng 4 2022
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).
tui chỉ làm dc này thui
13 tháng 11 2021
a) \(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)
c) đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+4\right)+2=t\left(t+3\right)+2=t^2+3t+2=\left(t^2+t\right)+\left(2t+2\right)=t\left(t+1\right)+2\left(t+1\right)=\left(t+1\right)\left(t+2\right)=\left(x^2+x+2\right)\left(x^2+x+3\right)\)
17 tháng 10 2021
a) \(=a\left(a^3-9a^2+a-9\right)=a\left[a^2\left(a-9\right)+\left(a-9\right)\right]\)
\(=a\left(a-9\right)\left(a^2+1\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(2y+z\right)+3\left(2y+z\right)=\left(2y+z\right)\left(x+3\right)\)
d) \(=x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-a\right)\left(x-b\right)\)
17 tháng 10 2021
a) = a(a³-9a²+a-9)
b) =3x²+5y-3xy-5x
= (3x²-5x)+(5y-3xy)
=x(3x-5)+y(5-3x)
=x(3x-5)-y(3x-5)
=(3x-5)(x-y)
c)2xy +3z+6y+xz
=(2xy+6y)+(3z+xz)
=2y(x+3)+z(3+x)
=(x+3)(2y-z)
NM
1
3 tháng 9 2021
a) \(x^2-9+2\left(x+3\right)=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x-3+2\right)=\left(x+3\right)\left(x-1\right)\)
b) \(x^2-10x+25-3\left(x-5\right)=\left(x-5\right)^2-3\left(x-5\right)=\left(x-5\right)\left(x-5-3\right)=\left(x-5\right)\left(x-8\right)\)
c) \(x^3-4x^2+3x=x\left(x^2-4x+3\right)=x\left(x-1\right)\left(x-3\right)\)
NL
1
8 tháng 10 2021
\(a,=5x\left(4x-1\right)\\ b,=y^2-\left(x-1\right)^2=\left(y-x+1\right)\left(y+x-1\right)\\ c,=6x^2+3x-4x-2=3x\left(x+2\right)-2\left(x+2\right)=\left(3x-2\right)\left(x+2\right)\)
18 tháng 1 2022
\(a,x\left(x+6\right)\\ b,\left(9x-1\right)\left(9x+1\right)\\ c,\left(x+y\right)-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ d,\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\)
Trả lời:
Ta có: \(A=\left(x^2+x\right)^2+5y\left(x^2+x\right)+6y^2\)
\(=\left(x^2+x\right)^2+2y\left(x^2+x\right)+3y\left(x^2+x\right)+6y^2\)
\(=\left[\left(x^2+x\right)^2+2y\left(x^2+x\right)\right]+\left[3y\left(x^2+x\right)+6y^2\right]\)
\(=\left(x^2+x\right)\left(x^2+x+2y\right)+3y\left(x^2+x+2y\right)\)
\(=\left(x^2+x+2y\right)\left(x^2+x+3y\right)\)