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14 tháng 8 2019

a)ta co:  125x^3+y^6=(5x)^3+(y^2)^3=(5x+y^2)(5x-5xy^2+y^2)                                                                                                                           b)ta co 5xy^2-10xyz+5xz^2=5x(y^2-2yz+z^2)=5x(y-z)^2                                                                                                                                                               (may cau sau gan giong ban tu lam nha)

15 tháng 8 2019

b) \(5xy^2-10xyz+5xz^2\)

\(=5xy^2-5xyz-5xyz+5xz^2\)

\(=5xy\left(y-z\right)-5xz\left(y-z\right)\)

\(=\left(y-z\right)\left(5xy-5xz\right)\)

\(=5x\left(y-z\right)\left(y-z\right)\)

\(=5x\left(y-z\right)^2\)

1 tháng 11 2016

Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...

a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)

\(=4\left[\left(x^2-2x+1\right)-\left(a^2-2ay+y^2\right)\right]\)

\(=4\left[\left(x-1\right)^2-\left(a-y\right)^2\right]\)

\(=4\left(x-1-a+y\right)\left(x-1+a-y\right)\)

b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)

\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)

c)\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x^2-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(5x+5\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)

\(=\left(x-1\right)\left(x^2+6x+9\right)\)

\(=\left(x-1\right)\left(x+3\right)^2\)

d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)

\(=\left(a+1\right)\left(a^4+a^2+1\right)\)

\(=\left(a+1\right)\left(a^4+2a^2+1-a^2\right)\)

\(=\left(a+1\right)\left[\left(a^2+1\right)^2-a^2\right]\)

\(=\left(a+1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)\)

e)\(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)

\(=\left(x-1-y\right)\left(x^2-2x+1+xy-y+y^2\right)\)

f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)

\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y\right)\left(5x-3y\right)\)

g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)

\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)

\(=3\left(a-b+c\right)\left(x+6y\right)^2\)

1 tháng 11 2016

a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)

\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)

\(=\left(2x-2\right)^2-\left(2y-2a\right)^2=\left(2x-2+2y-2a\right)\left(2x-2-2y+2a\right)\)

b/ \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)=\left(x+y-1\right)\left(x^2+y^2+2xy+x+y+1\right)-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)

Giải giúp bạn 2 bài tiêu biểu thôi nha

4 tháng 11 2016

b)(x2+x+1)(x2+x+2)-12

Đặt t=x2+x+1

t(t+1)-12=t2+t-12

=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)

=(x2+x-2)(x2+x+5)

=(x-1)(x+2)(x2+x+5)

c)(x2+8x+7)(x2+8x+15)+15

Đặt t=x2+8x+7 

t(t+8)+15=t2+8t+15

=(t+3)(t+5)

=(x2+8x+7+3)(x2+8x+7+15)

=(x2+8x+10)(x2+8x+22)

d)(x+2)(x+3)(x+4)(x+5)-24

=(x2+7x+10)(x2+7x+12)-24

Đặt t=x2+7x+10

t(t+2)-24=(t-4)(t+6)

=(x2+7x+10-4)(x2+7x+10+6)

=(x2+7x+6)(x2+7x+16)

=(x+1)(x+6)(x2+7x+16)

4 tháng 11 2016

a/ Đặt x2 + 4x + 8 = a

Thì đa thức ban đầu thành

a2 + 3ax + 2x= (a2 + 2ax + x2) + (ax + x2)

= (a + x)2 + x(a + x) = (a + x)(a + 2x)

Bài 1: Phân tích đa thức thành nhân tử: a) \(2x\left(x+1\right)+2\left(x+1\right)\) b) \(y^2\left(x^2+y\right)-zx^2-zy\) c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\) d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\) e) \(x^2-6xy+9y^2\) f) \(x^3+6x^2y+12xy^2+8y^3\) g) \(x^3-64\) h) \(125x^3+y^6\) k) \(0,125\left(a+1\right)^3-1\) t) \(x^2-2xy+y^2-xz+yz\) q) \(x^2-y^2-x+y\) p) \(a^3x-ab+b-x\) đ)...
Đọc tiếp

Bài 1: Phân tích đa thức thành nhân tử:

a) \(2x\left(x+1\right)+2\left(x+1\right)\)

b) \(y^2\left(x^2+y\right)-zx^2-zy\)

c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\)

d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

e) \(x^2-6xy+9y^2\)

f) \(x^3+6x^2y+12xy^2+8y^3\)

g) \(x^3-64\)

h) \(125x^3+y^6\)

k) \(0,125\left(a+1\right)^3-1\)

t) \(x^2-2xy+y^2-xz+yz\)

q) \(x^2-y^2-x+y\)

p) \(a^3x-ab+b-x\)

đ) \(3x^2\left(a+b+c\right)+36xy\left(a+b+c\right)+108y^2\left(a+b+c\right)\)

l) \(x^2-x-6\)

i) \(x^4+4x^2-5\)

m) \(x^3-19x-30\)

j) \(x^4+x+1\)

y) \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

o) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

ê) \(4a^2b^2-\left(a^2+b^2+c^2\right)^2\)

w) \(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)

z) \(\left(x^2-8\right)^2+36\)

u) \(81x^4+4\)

Bài 2 : Tìm x

a)\(\left(2x-1\right)^2-25=0\)

b) \(8x^3-50x=0\)

c) \(\left(x-2\right)\left(x^2+2+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

d) \(3x\left(x-1\right)+x-1=0\)

e) \(2\left(x+3\right)-x^2-3x\) =0

f) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

5
12 tháng 10 2017

Bài 1 :

a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)

b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)

c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)

d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)

e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)

12 tháng 10 2017

Bài 1 :

f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)

g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)

h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)

AH
Akai Haruma
Giáo viên
14 tháng 8 2019

a)

\(5x(x-2y)+2(2y-x)^2=5x(x-2y)+2(x-2y)^2\)

\(=(x-2y)[5x+2(x-2y)]=(x-2y)(7x-4y)\)

b)

\(7x(y-4)^2-(4-y)^3=7x(y-4)^2+(y-4)^3=(y-4)^2(7x+y-4)\)

c)

\((4x-8)(x^2+6)-(4x-8)(x+7)+9(8-4x)\)

\(=(4x-8)(x^2+6)-(4x-8)(x+7)-9(4x-8)\)

\(=(4x-8)[(x^2+6)-(x+7)-9]=(4x-8)(x^2-x-10)=4(x-2)(x^2-x-10)\)

d)

\(x^2-xz-9y^2+3yz=(x^2-9y^2)-(xz-3yz)\)

\(=(x-3y)(x+3y)-z(x-3y)=(x-3y)(x+3y-z)\)

e)

\(x^2(x^2-6)-x^2+9=x^4-7x^2+9=(x^4-6x^2+9)-x^2\)

\(=(x^2-3)^2-x^2=(x^2-3-x)(x^2-3+x)\)

AH
Akai Haruma
Giáo viên
12 tháng 8 2019

a)

\(5x(x-2y)+2(2y-x)^2=5x(x-2y)+2(x-2y)^2\)

\(=(x-2y)[5x+2(x-2y)]=(x-2y)(7x-4y)\)

b)

\(7x(y-4)^2-(4-y)^3=7x(y-4)^2+(y-4)^3=(y-4)^2(7x+y-4)\)

c)

\((4x-8)(x^2+6)-(4x-8)(x+7)+9(8-4x)\)

\(=(4x-8)(x^2+6)-(4x-8)(x+7)-9(4x-8)\)

\(=(4x-8)[(x^2+6)-(x+7)-9]=(4x-8)(x^2-x-10)=4(x-2)(x^2-x-10)\)

d)

\(x^2-xz-9y^2+3yz=(x^2-9y^2)-(xz-3yz)\)

\(=(x-3y)(x+3y)-z(x-3y)=(x-3y)(x+3y-z)\)

e)

\(x^2(x^2-6)-x^2+9=x^4-7x^2+9=(x^4-6x^2+9)-x^2\)

\(=(x^2-3)^2-x^2=(x^2-3-x)(x^2-3+x)\)