Phân tích đa thức sau thành nhân tử:
a) x3 + 6x2 + 11x + 6
b) x3 + 6x2 - 13x - 42
c) (x2 + x + 4)2 + 8x (x2 + x +4) + 15x2
d) (x + 2) (x + 4) (x + 6) (x + 8) + 16
Phân tích các đa thức sau thành nhân tử:
1) x3 - 7x + 6
2) x3 - 9x2 + 6x + 16
3) x3 - 6x2 - x + 30
4) 2x3 - x2 + 5x + 3
5) 27x3 - 27x2 + 18x - 4
6) x2 + 2xy + y2 - x - y - 12
7) (x + 2)(x +3)(x + 4)(x + 5) - 24
8) 4x4 - 32x2 + 1
9) 3(x4 + x2 + 1) - (x2 + x + 1)2
10) 64x4 + y4
11) a6 + a4 + a2b2 + b4 - b6
12) x3 + 3xy + y3 - 1
13) 4x4 + 4x3 + 5x2 + 2x + 1
14) x8 + x + 1
15) x8 + 3x4 + 4
16) 3x2 + 22xy + 11x + 37y + 7y2 +10
17) x4 - 8x + 63
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)
\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
7) \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\) (NHÂN x + 2 vs x + 5 và x + 3 vs x + 4 )
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
ĐẶT \(x^2+7x+11=y\) ta được :
\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)
\(=y^2-25=\left(y-5\right)\left(y+5\right)\)
8) \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)
\(=\left(2x^2+1\right)^2-\left(6x\right)^2\)\(=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
9) sai đề rùi bạn ơi ! đề đúng nè
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
Ta thấy :
\(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2\)\(=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
Thay vào biểu thức bài cho ta được :
\(3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)
\(=2\left(x^2+x+1\right)\left(x-1\right)^2\)
bài ở trên câu 3 : kết luận là \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs
Câu 6:Thực hiện phép nhân -2x(x2 + 3x - 4) ta được:
A.-2x3 - 6x2 – 8x B. 2x3 -6x2 – 8x C. -2x3 - 6x2 + 8x D. -2x3 + 3x2 -4
Câu 7 : Phân tích đa thức x2 + 2xy + y2 – 9z2 thành nhân tử ta được:
A. (x+y+3z)(x+y–3z)
B. (x-y+3z)(x+y–3z)
C.(x - y +3z)(x - y – 3z)
D. (x + y +3z)(x -y – 3z)
Câu 8: Phân tích đa thức 27x3 – thành nhân tử ta được:
A.(3x+)(9x2-x+)
B.(3x–)(9x2+x+)
C.(27x–)(9x2+x+)
D.(27x+)(9x2+x+)
Câu 9: Phân tích đa thức x2 + 7x + 12 thành nhân tử ta được:
A. (x - 3)( x + 4 ) B. (x + 3)( x + 4 ) C.(x + 5)( x + 2 ) D. (x -5)( x + 2 )
Câu 10: Giá trị của biểu thức (x2 + 4x + 4) tại x = - 2 là:
A. 4 B. -2 C. 0 D. -8
Câu 6:Thực hiện phép nhân -2x(x2 + 3x - 4) ta được:
A.-2x3 - 6x2 – 8x B. 2x3 -6x2 – 8x C. -2x3 - 6x2 + 8x D. -2x3 + 3x2 -4
Câu 7 : Phân tích đa thức x2 + 2xy + y2 – 9z2 thành nhân tử ta được:
A. (x+y+3z)(x+y–3z)
B. (x-y+3z)(x+y–3z)
C.(x - y +3z)(x - y – 3z)
D. (x + y +3z)(x -y – 3z)
Câu 9: Phân tích đa thức x2 + 7x + 12 thành nhân tử ta được:
A. (x - 3)( x + 4 ) B. (x + 3)( x + 4 ) C.(x + 5)( x + 2 ) D. (x -5)( x + 2 )
Câu 10: Giá trị của biểu thức (x2 + 4x + 4) tại x = - 2 là:
A. 4 B. -2 C. 0 D. -8
Mấy câu còn lại bị lỗi r nhé
Phân tích các đa thức sau thành nhân tử:
1) x3 - 7x + 6
2) x3 - 9x2 + 6x + 16
3) x3 - 6x2 - x + 30
4) 2x3 - x2 + 5x + 3
5) 27x3 - 27x2 + 18x - 4
6) x2 + 2xy + y2 - x - y - 12
7) (x + 2)(x +3)(x + 4)(x + 5) - 24
8) 4x4 - 32x2 + 1
9) 3(x4 + x2 + 1) - (x2 + x + 1)2
10) 64x4 + y4
11) a6 + a4 + a2b2 + b4 - b6
12) x3 + 3xy + y3 - 1
13) 4x4 + 4x3 + 5x2 + 2x + 1
14) x8 + x + 1
15) x8 + 3x4 + 4
16) 3x2 + 22xy + 11x + 37y + 7y2 +10
17) x4 - 8x + 63
Chia nhỏ ra cậu ơi :v
Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((
Phân tích các đa thức sau thành nhân tử:
1) x3 - 7x + 6
2) x3 - 9x2 + 6x + 16
3) x3 - 6x2 - x + 30
4) 2x3 - x2 + 5x + 3
5) 27x3 - 27x2 + 18x - 4
`1)x^3-7x+6`
`=x^3-x-6x+6`
`=x(x-1)(x+1)-6(x-1)`
`=(x-1)(x^2+x-6)`
`=(x-1)(x^2-2x+3x-6)`
`=(x-1)[x(x-2)+3(x-2)]`
`=(x-1)(x-2)(x+3)`
`2)x^3-9x^2+6x+16`
`=x^3-2x^2-7x^2+14x-8x+16`
`=x^2(x-2)-7x(x-2)-8(x-2)`
`=(x-2)(x^2-7x-8)`
`=(x-2)(x^2-8x+x-8)`
`=(x-2)[x(x-8)+x-8]`
`=(x-2)(x-8)(x+1)`
`3)x^3-6x^2-x+30`
`=x^3+2x^2-8x^2-16x+15x+30`
`=x^2(x+2)-8x(x+2)+15(x+2)`
`=(x+2)(x^2-8x+15)`
`=(x+2)(x^2-3x-5x+15)`
`=(x+2)[x(x-3)-5(x-3)]`
`=(x+2)(x-3)(x-5)`
`4)2x^3-x^2+5x+3`
`=2x^3+x^2-2x^2-x+6x+3`
`=x^2(2x+1)-x(2x+1)+3(2x+1)`
`=(2x+1)(x^2-x+3)`
`5)27x^3-27x^2+18x-4`
`=27x^3-9x^2-18x^2+6x+12x-4`
`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`
`=(3x-1)(9x^2-6x+4)`
1) Ta có: \(x^3-7x+6\)
\(=x^3-x-6x+6\)
\(=x\left(x^2-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
2) Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
3) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
4) Ta có: \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+6\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+6\right)\)
5) Ta có: \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
Phân tích các đa thức sau thành nhân tử:
1) x3 - 7x + 6
2) x3 - 9x2 + 6x + 16
3) x3 - 6x2 - x + 30
4) 2x3 - x2 + 5x + 3
5) 27x3 - 27x2 + 18x - 4
6) x2 + 2xy + y2 - x - y - 12
7) (x + 2)(x +3)(x + 4)(x + 5) - 24
8) 4x4 - 32x2 + 1
9) 3(x4 + x2 + 1) - (x2 + x + 1)2
10) 64x4 + y4
11) a6 + a4 + a2b2 + b4 - b6
12) x3 + 3xy + y3 - 1
13) 4x4 + 4x3 + 5x2 + 2x + 1
14) x8 + x + 1
15) x8 + 3x4 + 4
16) 3x2 + 22xy + 11x + 37y + 7y2 +10
17) x4 - 8x + 63
a,\(x^3-7x+6\)
\(=x^3-2x^2+2x^2-4x-3x+6\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)
\(=x^2.\left(x-2\right)+2x.\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)
\(=\left(x-2\right).\left[x.\left(x-1\right)+3.\left(x-1\right)\right]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
b,\(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)
\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)
\(=\left(x-8\right).\left(x^2-x-2\right)\)
\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)
\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)
\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)
\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)
c,\(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2.\left(x-5\right)-x.\left(x-5\right)-6.\left(x-5\right)\)
\(=\left(x-5\right).\left(x^2-x-6\right)\)
\(=\left(x-5\right).\left(x^2+2x-3x-6\right)\)
\(=\left(x-5\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)
\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)
\(=\left(x-5\right).\left(x+2\right).\left(x-3\right)\)
Chúc bạn học tốt!!!
d,\(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2.\left(2x+1\right)-x.\left(2x+1\right)+3.\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x^2-x+3\right)\)
e, \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)
\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)
Chúc bạn học tốt!!!
7, \(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)
\(=\left[\left(x+2\right).\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)
\(=\left(x^2+5x+2x+10\right).\left(x^2+4x+3x+12\right)-24\)
\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)(1)
Đặt \(t=x^2+7x+10\Rightarrow t+2=x^2+7x+12\)
\(\Rightarrow\left(1\right)=t.\left(t+2\right)-24\)
\(=t^2+2t-24=t^2-4t+6t-24\)
\(=\left(t^2-4t\right)+\left(6t-24\right)=t.\left(t-4\right)+6.\left(t-4\right)\)
\(=\left(t-4\right).\left(t+6\right)\) (2)
Vì \(t=x^2+7x+10\) nên:
(2) \(=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right).\left(x^2+7x+16\right)\)
\(=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)
\(=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)
\(=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)
Chúc bạn học tốt!!!
Bài 2: Phân tích các đa thức sau thành nhân tử bằng phương pháp dùng hằng đẳng thức
a)x2-4x+4 b)4x2+4x+1 c)16x2-9y2
d)16-(x+3)2 e)4x2-(3x-1)2 f)x3-y3
g)27+x3 h)x3+6x2+12x+8 i)1-3x+3x2-x3
giúp mình cần gấp ,mn ơi
a) \(=\left(x-2\right)^2\)
b) \(=\left(2x+1\right)^2\)
c) \(=\left(4x-3y\right)\left(4x+3y\right)\)
d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)
e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)
f) \(=\left(x-y\right)\left(x^2+xy+y^2\right)\)
g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)
h) \(=\left(x+2\right)^3\)
i) \(=\left(1-x\right)^3\)
a/ $=(x-2)^2$
b/ $=(2x+1)^2$
c/ $=(4x-3y)(4x+3y)$
d/ $=(1-x)(x+7)$
e/ $=(-x+1)(5x-1)$
f/ $=(x-y)(x^2+xy+y^2)$
g/ $=(3+x)(9-3x+x^2)$
h/ $=(x+2)^3$
i/ $=(1-x)^3$
Bài 2: Phân tích các đa thức sau thành nhân tử bằng phương pháp dùng hằng đẳng thức
a)x2-4x+4 b)4x2+4x+1 c)16x2-9y2
d)16-(x+3)2 e)4x2-(3x-1)2 f)x3-y3
g)27+x3 h)x3+6x2+12x+8 i)1-3x+3x2-x3
giúp mình cần gấp ,mn ơi
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(4x^2+4x+1=\left(2x+1\right)^2\)
g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
Bài 1: Phân tích đa thức thành nhân tử: a) 4y3 + 16y2 + 16y b) 8x2-48x+6xy-36y c) 8x2-48x-6xy+36y d) a2 –2ab+b2 –4 e) 4–x2 –4xy–4y2 f) 8a2 –16a+8ax–16x g) 16–4x2 +8xy–4y2 h) –4x2 –16xy–16y2 Bài 2: Tìm x, biết: a) x3 – 6x2 + 9x = 0 b) 5x(x–6)+3x–18=0 c) 5x(x – 6) – 18 + 3x = 0 d) 5x(x – 6) – 3x + 18 = 0 e) (2x – 3)2 = (5 – x)2 f) (2x + 1)2 = (3x – 2)2 g) 16(2x–3)=-25x2 (3–2x)
b: \(8x^2-48x+6xy-36y\)
\(=8x\left(x-6\right)+6y\left(x-6\right)\)
\(=2\left(x-6\right)\left(4x+3y\right)\)
d: \(a^2-2ab+b^2-4\)
\(=\left(a-b\right)^2-4\)
\(=\left(a-b-2\right)\left(a-b+2\right)\)
Bài 1:Phân tích đa thức thành nhân tử:
a) x3y+x-y-1
b) x2.(x-2)+4.(2-x)
c) x3-x2-20x
d) (x2+1)2-(x+1)2
e) 6x2-7x+2
f) x4+8x2+12
g) (x3+x+1).(x3+x)-2
h) (x+1).(x+2).(x+3).(x+4)-1
i) -(x2+2)2+4x.(x2+2)-3x2
j) -(x2+2)2+4x.(x2+2).3x2
k) -(x2+2)2+4x.(x2+2)+3x2
l) 81x4+4y4
Giúp với ạa
a) x³y + x - y - 1
= (x³y - y) + (x - 1)
= y(x³ - 1) + (x - 1)
= y(x - 1)(x² + x + 1) + (x - 1)
= (x - 1)[y(x² + x + 1) + 1]
= (x - 1)(x²y + xy + y + 1)
b) x²(x - 2) + 4(2 - x)
= x²(x - 2) - 4(x - 2)
= (x - 2)(x² - 4)
= (x - 2)(x - 2)(x + 2)
= (x - 2)²(x + 2)
c) x³ - x² - 20x
= x(x² - x - 20)
= x(x² + 4x - 5x - 20)
= x[(x² + 4x) - (5x + 20)]
= x[x(x + 4) - 5(x + 4)]
= x(x + 4)(x - 5)
d) (x² + 1)² - (x + 1)²
= (x² + 1 - x - 1)(x² + 1 + x + 1)
= (x² - x)(x² + x + 2)
= x(x - 1)(x² + x + 2)
e) 6x² - 7x + 2
= 6x² - 3x - 4x + 2
= (6x² - 3x) - (4x - 2)
= 3x(2x - 1) - 2(2x - 1)
= (2x - 1)(3x - 2)
f) x⁴ + 8x² + 12
= x⁴ + 2x² + 6x² + 12
= (x⁴ + 2x²) + (6x² + 12)
= x²(x² + 2) + 6(x² + 2)
= (x² + 2)(x² + 6)
g) (x³ + x + 1)(x³ + x) - 2
Đặt u = x³ + x
x³ + x + 1 = u + 1
(u + 1).u - 2
= u² + u - 2
= u² - u + 2u - 2
= (u² - u) + (2u - 2)
= u(u - 1) + 2(u - 1)
= (u - 1)(u + 2)
= (x³ + x - 1)(x³ + x + 2)
= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)
= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]
= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]
= (x³ + x - 1)(x - 1)(x² - x + 2)
h) (x + 1)(x + 2)(x + 3)(x + 4) - 1
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1
= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)
Đặt u = x² + 5x + 4
u + 2 = x² + 5x + 6
(1) u.(u + 2) - 1
= u² + 2u - 1
= u² + 2u + 1 - 2
= (u² + 2u + 1) - 2
= (u + 1)² - 2
= (u + 1 + √2)(u + 1 - √2)
= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)
= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)
i: \(-\left(x^2+2\right)^2+4x\left(x^2+2\right)-3x^2\)
\(=-\left[\left(x^2+2\right)^2-4x\left(x^2+2\right)+3x^2\right]\)
\(=-\left[\left(x^2+2\right)^2-x\left(x^2+2\right)-3x\left(x^2+2\right)+3x^2\right]\)
\(=-\left[\left(x^2+2\right)\left(x^2+2-x\right)-3x\left(x^2+2-x\right)\right]\)
\(=-\left(x^2+2-x\right)\left(x^2-3x+2\right)\)
\(=-\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x-1\right)\)
\(=-\left(x+2\right)\left(x-2\right)\left(x-1\right)^2\)
l: \(81x^4+4y^4\)
\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)
\(=\left(81x^4+36x^2y^2+4y^4\right)-\left(6xy\right)^2\)
\(=\left[\left(9x^2\right)^2+2\cdot9x^2\cdot2y^2+\left(2y^2\right)^2\right]-\left(6xy\right)^2\)
\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)
\(=\left(9x^2+2y^2+6xy\right)\left(9x^2+2y^2-6xy\right)\)
Bài 3: Phân tích các đa thức sau thành nhân tử:
a) x2 + 10x + 25. b) 8x - 16 - x2
c) x3 + 3x2 + 3x + 1 d) (x + y)2 - 9x2
e) (x + 5)2 – (2x -1)2
Bài 4: Tìm x biết
a) x2 – 9 = 0 b) (x – 4)2 – 36 = 0
c) x2 – 10x = -25 d) x2 + 5x + 6 = 0
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2