phân tích đa thức thành nhân tử
a ) x2 - x - 12
b) x8 + 3x4 + 4
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TM
11 tháng 8 2021
a/ \(\left(x+y\right)^2-8\left(x+y\right)+12\)
\(=\left(x+y\right)\left(x+y-8+12\right)\)
\(=\left(x+y\right)\left(x+y+4\right)\)
==========
b/\(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-\left(2x^2+4x\right)-3\)
\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3\)
\(=\left(x^2+2x\right)\left(x^2+2x-5\right)\)
===========
c/ \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2-15\right)\)
\(=\left(x^2+x\right)\left(x^2+x-17\right)\)
[---]
KV
25 tháng 5 2023
x⁸ + x⁴ + 1
= x⁸ + 2x⁴ + 1 - x⁴
= (x⁴ + 1)² - x⁴
= (x⁴ + 1)² - (x²)²
= (x⁴ + 1 + x²)(x⁴ + 1 - x²)
= (x⁴ + x² + 1)(x⁴ - x² + 1)
AH
Akai Haruma
Giáo viên
7 tháng 9 2021
Lời giải:
a.
$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$
$=(x^4+1-x^2)[(x^2+1)^2-x^2]$
$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$
b.
$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$
$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$
c.
$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$
$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$
$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$
c.
$x^2-5y^2-y^4+2xy-9$
$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$
7 tháng 9 2021
\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)
22 tháng 7 2023
a: (x^2+x)^2+4x^2+4x-12
=(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
b: =(x^2+8x)^2+22(x^2+8x)+105+15
=(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
c: =8x^2+12x-2x-3
=(2x+3)(4x-1)
23 tháng 7 2023
a: =(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
b: =(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+12)(x^2+8x+10)
=(x+2)(x+6)(x^2+8x+10)
c: =8x^2+12x-2x-3
=(2x+3)(4x-1)
NH
2
25 tháng 8 2021
a) \(x^2\left(x^2+4\right)-x^2-4=x^2\left(x^2+4\right)-\left(x^2+4\right)=\left(x^2+4\right)\left(x^2-1\right)=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
NT
25 tháng 8 2021
a. \(x^2\left(x^2+4\right)-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2-1\right)\left(x^2+4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+4\right)\)
b. \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)
Đặt \(t=x^2+7x+10\), ta được
(*) \(=t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)\)
hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)
15 tháng 9 2021
\(a,=\left(x-1\right)^4-2\left(x-1\right)^2+1\\ =\left[\left(x-1\right)^2-1\right]^2\\ =\left(x^2-2x-2\right)^2\\ b,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-4\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36\\ =\left(x^2+6x+4\right)\left(x^2+6x+9\right)\\ =\left(x+3\right)^2\left(x^2+6x+4\right)\)
27 tháng 9 2021
a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)
c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
27 tháng 9 2021
a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=x^4-14x^2-32\)
\(=\left(x^2-16\right)\left(x^2+2\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)
\(=\left(x^2+5x+1\right)^2\)
a) ta có : x^2 -x-12 =( x^2 -4x) +(3x-12)=x(x-4) + 3(x-4) =(x+3)(x-4)
b)ta có: x^8 +3x^4 -4= x^4(x^4 +4) - (x^4 +4) =( x^4 -1)(x^4 +4) =(x^2 -1)(x^2 +1)(x^4 +4)
\(x^8+3x^4+4\)
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4+2\right)^2-x^4\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)