phân tích các đa thức sau thành nhân tử bằng phương pháp tách một hạng tử thành 2
x2 -6x +8
3x2- 8x + 4
9x2 +6x - 8
4x2-4x-3
x2- 7x +12
x2- 5x - 14
PHÂN TÍCH CÁC ĐA THỨC SAU THÀNH NHÂN TỬ BẰNG PHƯƠNG PHÁP NHÓM NHIỀU HẠNG TỬ :
a) x2 -2x -4y2-4y
b) x4 + 2x3 - 4x -4
c) x3 + 2x2y -x -2y
d) 3x2 -3y2 -2(x-y)2
e) x3 -4x2 -9x +36
f) x2 -y2 -2x -2y
a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
BT3: Phân tích các đa thức sau thành nhân tử bằng phương pháp cách tách hạng tử. a, x^3 + 4x^2 - 21x b, 5x^3 + 6x^2 + x c, x^3 - 7x + 6 d, 3x^3 + 2x - 5
a) \(x^3+4x^2-21x\)
\(=x\left(x^2+4x-21\right)\)
\(=x\left(x^2-3x+7x-21\right)\)
\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)
\(=x\left(x-3\right)\left(x+7\right)\)
b) \(5x^3+6x^2+x\)
\(=x\left(5x^2+6x+1\right)\)
\(=x\left(5x^2+5x+x+1\right)\)
\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(5x+1\right)\)
c) \(x^3-7x+6\)
\(=x^3+2x^2-3x-2x^2-4x+6\)
\(=x\left(x^2+2x-3\right)-2\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+3\right)\)
d) \(3x^3+2x-5\)
\(=3x^3+3x^2+5x-3x^2-3x-5\)
\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)
\(=\left(x-1\right)\left(3x^2+3x+5\right)\)
phân tích đa thức thành nhân tử bằng phương pháp tách hạng tử
a 4x^3 - 13 x^2 + 9x - 18
b - x^3 - 6x^2 + 6x +1
c x^3 - 4x^2 - 8x + 8
a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)
b. \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)
c. \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)
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
`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 đa thức thành nhân tử bằng phương pháp tách hạng tử
x^3 - 5x^2 + 8x - 4
x^3 - 9x^2 + 6x +16
a, = (x^3-x^2)-(4x^2-4x)+(4x-4)
= (x-1).(x^2-4x+4) = (x-1).(x-2)^2
b, = (x^3+x^2)-(10x^2+10x)+(16x+16)
= (x+1).(x^2-10x+16)
= (x+1).[ (x^2-2x)-(8x-16) ] = (x+1).(x-2).(x-8)
k mk nha
a)= (x^3-x^2)-(4x^2-4x)+(4x-4)
= (x-1).(x^2-4x+4)
= (x-1).(x-2)^2
b)= (x^3+x^2)-(10x^2+10x)+(16x+16)
= (x+1).(x^2-10x+16)
= (x+1).[ (x^2-2x)-(8x-16) ]
= (x+1).(x-2).(x-8)
P/s tham khảo nha
Bài 2 Phân tích thành nhân tử
a) 3x2 – 7x – 10
b) x2 + 6x +9 – 4y2
c) x2 – 2xy + y2 – 5x + 5y’
d) 4x2 – y2 – 6x + 3y
e) 1 – 2a + 2bc + a2 – b2 – c2
f) x3 – 3x2 – 4x + 12
g) x4 + 64
h) x4 – 5x2 + 4
i) (x+1)(x+3)(x+5)(x+7) + 16
j) (x2 + 6x +8)( x2 + 14x + 48) – 9
k) ( x2 – 8x + 15)(x2 – 16x + 60) – 24x2
l) 4( x2 + 15x + 50)(x2 +18x +72) – 3x2
Bài 3 tìm gtnn
A = 9x2 – 6x + 2
B = 4x2 + 5x + 10
C = x2 – x + 10
D = 4x2 + 3x + 20
E = x2 + y2 – 6xy + 10y + 35
F= x2 + y2 – 6x + 4y +2
M= 2x2 + 4y2 – 4xy – 4x – 4y +2021
Bài 2:
a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)
b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)
d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)
g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)
i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)
a: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=\left(x+1\right)\left(3x-10\right)\)
b: \(x^2+6x+9-4y^2\)
\(=\left(x+3\right)^2-4y^2\)
\(=\left(x+3-2y\right)\left(x+3+2y\right)\)
c: \(x^2-2xy+y^2-5x+5y\)
\(=\left(x-y\right)^2-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-5\right)\)
a) 3x2−7x−10=(x+1)(3x−10)3x2−7x−10=(x+1)(3x−10)
b) x2+6x+9−4y2=(x+3)2−(2y)2=(x+3−2y)(x+3+2y)x2+6x+9−4y2=(x+3)2−(2y)2=(x+3−2y)(x+3+2y)
c) x2−2xy+y2−5x+5y=(x−y)2−5(x−y)=(x−y)(x−y−5)x2−2xy+y2−5x+5y=(x−y)2−5(x−y)=(x−y)(x−y−5)
d) 4x2−y2−6x+3y=(2x−y)(2x+y)−3(2x−y)=(2x−y)(2x+y−3)4x2−y2−6x+3y=(2x−y)(2x+y)−3(2x−y)=(2x−y)(2x+y−3)
e) 1−2a+2bc+a2−b2−c2=(a−1)2−(b−c)2=(a−1−b+c)(a−1+b−c)1−2a+2bc+a2−b2−c2=(a−1)2−(b−c)2=(a−1−b+c)(a−1+b−c)
f) x3−3x2−4x+12=(x+2)(x−3)(x−2)x3−3x2−4x+12=(x+2)(x−3)(x−2)
g) x4+64=(x2+8)2−16x2=(x2+8−4x)(x2+6+4x)x4+64=(x2+8)2−16x2=(x2+8−4x)(x2+6+4x)h) x4−5x2+4=(x+2)(x+1)(x−1)(x−2)x4−5x2+4=(x+2)(x+1)(x−1)(x−2)
i) (x+1)(x+3)(x+5)(x+7)+16=(x2+8x+7)(x2+8x+15)+16=(x2+8x+7)2+8(x2+8x+7)+16=(x2+8x+11)2(x+1)(x+3)(x+5)(x+7)+16=(x2+8x+7)(x2+8x+15)+16=(x2+8x+7)2+8(x2+8x+7)+16=(x2+8x+11)2
phân tích đa thức thành nhân tử
a/ x2 - 4x + 4 – y2 e/ 25x2 - 4y2
b/ 4x4 + 8x3 + 4x2 f/ x2 + 7x + 12
c/ x3y2 – 2x2y3 + xy4 i/ x2 - 5x - 14
d/ x2 - y2 – 7x + 7y
giúp mình với mình đang cần gấp ạ
\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)
Phân tích các đa thức sau thành nhân tử bằng phương pháp tách 1 hạng tử thành 2:
a)9x^2+6x-8
b)4x^2-4x-3
Giup mik vs mik cần gấp
Lời nói chẳng mất tiền mua. Lựa lời mà chửi cho vừa lòng nhau. Đã chửi, phải chửi thật đau. Chửi mà hiền quá còn lâu nó chừa. Chửi đúng , không được chửi bừa . Chửi cha mẹ nó , không thừa một ai . Khi chửi , chửi lớn mới oai. Chửi hay là phải chửi dài , chửi lâu . Chửi đi chửi lại mới ngầu. Chửi nhiều cho nó nhức đầu , đau tai. Chửi xong nhớ nói bái bai . Phóng nhanh kẻo bị ăn chai vào mồm.
\(a,9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^3\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)=\left(3x-2\right)\left(3x+4\right)\)
\(b,4x^2-4x-3\)
\(=\left(2x\right)^2-2.2x+1-4\)
\(=\left(2x-1\right)^2-2^2\)
\(=\left(2x-1-2\right)\left(2x-1+2\right)=\left(2x-3\right)\left(2x+1\right)\)
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
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