Phân tích các đa thức sau thành nhân tử:
A) \(2x^2+7x+5\)
B) \(x^2-4x-5\)
C) \(x^4+x^3+x+1\)
D) \(x^4+4x^2-5\)
Mình đang cần lời giải (chi tiết). Cảm ơn
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a: =(16x+20)^2-(10x+10)^2
=(16x+20-10x-10)(16x+20+10x+10)
=(26x+30)(6x+10)
=4(13x+15)(3x+5)
b: =(x-y+4-2x-3y+1)(x-y+4+2x+3y-1)
=(-x-4y+5)(3x+2y+3)
c: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]
=(x^2+2x+1-x^2+2x-1)(x^2+2x+1+x^2-2x+1)
=2(x^2+1)*4x
=8x(x^2+1)
Thứ nhất em làm quá tắt, thứ 2 em trình bày nó rất là khó nhìn. Em làm nhanh cho có số lượng chứ anh thấy làm thế sao mấy bạn hỏi bài hiểu được hả em? Làm bằng cái tâm nha em!
a) = (x + 1)^3 - 27z^3 = (x+1 - 3z)( (x+1)^2 + 3z(x+1) + 9z^2 )
b)= x^2 + x+ 3x + 3 = x (x+1) +3 (x+1) =(x+3)(x+1)
c) = 2x^2 - 2x + 5x - 5 = 2x(x-1) + 5(x-1) = (2x+5)(x-1)
d) = (a^2 + 1 - 2a)(a^2 +2a +1) = (a-1)^2 * (a+1)^2
e) = x^3 ( x-1) - (x^2 - 1) = x^3 ( x-1) - (x+1)(x-1) = (x^3 -x -1)(x-1)
a) Sửa đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
b) \(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,Sửa:x^2-2x+2y-y^2=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\\ d,=\left(4x^4+36x^2+81\right)-36x^2\\ =\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\\ e,=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x^2+x-x+1\\ =x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) x3 -2x2 +5x-4
=x3-x2-x2+x+4x-4
=x2(x-1)-x(x-1)+4(x-1)
=(x2-x+4)(x-1)
b) x3-x2+x+3
=x3+x2-2x2-2x+3x+3
=x2(x+1) -2x(x+1)+3(x+1)
=(x2-2x+3)(x+1)
c) 6x3+x2+x+1
=6x3+ 3x2-2x2-x+2x+1
=6x2(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))
=(6x2-2x+2) (x+\(\frac{1}{2}\))
=2( 3x2-x+1) (x+\(\frac{1}{2}\))
d) 4x3 + 6x2+4x+1
= 4x3+2x2+4x2+2x+2x+1
= 4x2(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))
= 2(2x2 +2x+1)( x+\(\frac{1}{2}\))
e) x6 -9x3+8
A/ \(16x-5x^2-3=\left(15x-3\right)-\left(5x^2-x\right)=3\left(5x-1\right)-x\left(5x-1\right)=\left(5x-1\right)\left(3-x\right)\)
B/ \(x^3-3x^2+1-3x=\left(x^3-4x^2+x\right)+\left(x^2-4x+1\right)=x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
C/ \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
D/ \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
`#\text {Kr.Ryo}`
`a)`
`4x^2 - 4x + 1`
`= (2x)^2 - 2*2x*1 + 1^2`
`= (2x - 1)^2`
`b)`
Xem lại đề
`c)`
`2x^2 + 7x + 5`
`= 2x^2 + 2x + 5x + 5`
`= (2x^2 + 2x) + (5x + 5)`
`= 2x(x + 1) + 5(x + 1)`
`= (2x + 5)(x + 1)`
`d)`
`x^2 - 6xy - 25z^2 + 9y^2`
`= (x^2 - 6xy + 9y^2) - 25z^2`
`= [ (x)^2 - 2*x*3y + (3y)^2] - (5z)^2`
`= (x + 3y)^2 - (5z)^2`
`= (x + 3y - 5z)(x + 3y + 5z)`
\(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)
\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)
\(+\left(x^7-x^5+x^4-x^2+x\right)\)
\(+\left(x^6-x^4+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
A/ \(2x^2+7x+5=2\left(x^2+2x+1\right)+3x+3=2\left(x+1\right)^2+3\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
B/ \(x^2-4x-5=\left(x^2-4x+4\right)-9=\left(x-2\right)^2-3^2=\left(x-5\right)\left(x+1\right)\)
C/ \(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
D/\(x^4+4x^2-5=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-3^2=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) = 2x^2 + 2x +5x + 5 = 2x(x+1) + 5(x+1) = (2x+5)(x+1)
b) = x^2 + x - 5x - 5 = x(x-1) - 5(x-1) = (x-5)(x-1)
c) = x^3 ( x+1) + x+1 = (x^3+1) (x+1) = (x+1)^2 * (x^2 - x +1)
d) = x^4 - x^2 + 5x^2 -5 = x^2 (x^2-1) + 5(x^2-1) = (x^2+5)(x-1)(x+1)