giải pt nghiệm nguyên : x4+2x3+2x2+2x+1=y2
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18 tháng 7 2021
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
18 tháng 7 2021
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
31 tháng 8 2021
a: Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)\)
b: Ta có: \(-a^4+a^3+2a^3+2a^2\)
\(=-a^2\left(a^2-a-2a-2\right)\)
c: Ta có: \(x^4+x^3+2x^2+x+1\)
\(=x^4+x^3+x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^2+1\right)\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
TG
18 tháng 7 2021
a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
b) \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)
d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)
18 tháng 7 2021
a) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b) Ta có: \(x^3+2x^2+2x+1\)
\(=\left(x^3+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
9 tháng 12 2021
a) 2x. (x2 – 7x -3)
= 2x3- 14x2- 6x
b) ( -2x3 + y2 -7xy). 4xy2
= -8x4y2+ 4xy4- 28x2y3
c)(-5x3).(2x2+3x-5)
= -10x5-15x4+25x3
d) (2x2 - xy+ y2).(-3x3)
=-6x5+ 3x4y -3x3y2
e)(x2 -2x+3). (x-4)
=x3-2x2+3x -4x2+8x-12
=x3-6x2+11x-12
f) ( 2x3 -3x -1). (5x+2)
=10x4-15x2-5x +4x3-6x-2
=10x4+4x3-15x2-11x-2
CM
25 tháng 12 2018
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].