\(^2\) - y\(^2\)
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Bài 1: Phân tích đa thức thành nhân tử

a) Ta có: \(16x^2-y^2+6y-9\)

\(=16x^2-\left(y^2-6y+9\right)\)

\(=\left(4x\right)^2-\left(y-3\right)^2\)

\(=\left[4x-\left(y-3\right)\right]\left[4x+\left(y-3\right)\right]\)

\(=\left(4x-y+3\right)\left(4x+y-3\right)\)

b) Ta có: \(a^2-16a^2b^2+b^2+2ab\)

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

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

\(=\left(a+b-4ab\right)\left(a+b+4ab\right)\)

c) Ta có: \(x^3-6x^2-9x\)

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

d) Ta có: \(mx^2+my^2-nx^2-ny^2\)

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

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

e) Ta có: \(a^3+b^3+a^2c+b^2c-abc\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)

\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)\)

f) Ta có: \(4x^2-y^2-4x+1\)

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

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

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

g) Ta có: \(\left(2x+3\right)^2+5\cdot\left(2x+3\right)\)

\(=\left(2x+3\right)\left(2x+3+5\right)\)

\(=\left(2x+3\right)\left(2x+8\right)\)

\(=2\left(2x+3\right)\left(x+4\right)\)

h) Ta có: \(3x^2-10x-8\)

\(=3x^2-12x+2x-8\)

\(=3x\left(x-4\right)+2\left(x-4\right)\)

\(=\left(x-4\right)\left(3x+2\right)\)

18 tháng 7 2020

cảm ơn nhiều ạ :)))

1 tháng 10 2018

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

=x.(x2-2x+1)

=x(x-1)2

B=\(2x^2+4x+2-2y^2\)

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

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

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

C=\(2xy-x^2-y^2+16\)

=\(-\left(-2xy+x^2+y^2-16\right)\)

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

=-(x-y-4)(x-y+4)

D=\(x^3+2x^2y+xy^2-9x\)

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

=\(x.\left[\left(x-y\right)^2-3^2\right]\)

=x.(x-y-3)(x-y+3)

E=\(2x-2y-x^2+2xy-y^2\)

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

=\(2\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)

=(x-y)(2x-2y-x+y)

=(x-y)(x+y)

1 tháng 10 2018

ở câu B:

(x+1)^1 sửa giùm mk thành (x+1)^2

4 tháng 12 2018

a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)

\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)

\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)

\(=-2x^2+2x+6\)

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

b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)

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

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

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

\(=x^4+4x^2+4-x^4+16\)

\(=4x^2+20\)

\(=4\left(x^2+5\right)\)

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

\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)

\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)

\(=-7x^2-20xy-17y^2+1\)

d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)

\(=x^6-3x^4+3x^2-1-x^6+1\)

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

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

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

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

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

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

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

\(=\left(-2\right)^2=4\)

g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)

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

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

\(=\left(x+2z\right)^2\)

h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)

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

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

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

\(=\left(-2\right)^2=4\)

i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)

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

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

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

\(=7x^2-4\)

Bài 1: Phân tích đa thức thành nhân tử

a) Ta có: \(8a^3-6a^2-1+3a\)

\(=\left[\left(2a\right)^3-1^3\right]-3a\left(2a-1\right)\)

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

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

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

b) Ta có: \(x^3-2x^2y+xy^2-9x\)

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

\(=x\left[\left(x^2-2xy+y^2\right)-9\right]\)

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

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

c) Ta có: \(5x^2-45\)

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

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

d) Ta có: \(2x^3-4x^2+2x\)

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

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

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

\(=x\left(x-1\right)\left(2x-2\right)\)

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

e) Ta có: \(6x\left(3x-2\right)-12\left(2-3x\right)\)

\(=6x\left(3x-2\right)+12\left(3x-2\right)\)

\(=\left(3x-2\right)\left(6x+12\right)\)

\(=6\left(3x-2\right)\left(x+2\right)\)

f) Ta có: \(4x^2-8xy+4y^2-10\)

\(=\left(2x\right)^2-2\cdot2x\cdot2y+\left(2y\right)^2-10\)

\(=\left(2x-2y\right)^2-10\)

\(=\left(2x-2y-\sqrt{10}\right)\left(2x-2y+\sqrt{10}\right)\)

g) Ta có: \(2x^2-8x+8\)

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

\(=2\left(x-2\right)^2\)

h) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

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

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

\(=3x\left(x+2\right)\)

13 tháng 10 2019

\(e,-5x+x^2-14\)

\(=x^2+2x-7x-14\)

\(=x\left(x+2\right)-7\left(x+2\right)\)

\(=\left(x+2\right)\left(x-7\right)\)

\(f,x^3+8+6x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)

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

\(g,15x^2-7xy-2y^2\)

\(=15x^2+3xy-10xy-2y^2\)

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

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

\(h,3x^2-16x+5\)

\(=3x^2-x-15x+5\)

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

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

13 tháng 10 2019

\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)

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

\(b,4x^2-9y^2+4x-6y\)

\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)

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

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

\(c,-x^2+5x+2xy-5y-y^2\)

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

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

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

\(d,x^2+4x-12\)

\(=x^2-2x+6x-12\)

\(=x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

22 tháng 8 2018

xin lỗi bài này mình không biết

24 tháng 9 2017

Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@

B1: Phân tích thành nhân tử:

a) \(6x^2+9x=3x\left(2x+3\right)\)

b) \(4x^2+8x=4x\left(x+2\right)\)

c) \(5x^2+10x=5x\left(x+2\right)\)

d) \(2x^2-8x=2x\left(x-4\right)\)

e) \(5x-15y=5\left(x-3y\right)\)

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

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

g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)

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

h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)

i) \(9x^2-18x+9=\left(3x-3\right)^2\)

k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)

m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)

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

n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)

\(=\left(x-31\right)\left(x+1\right)\)

o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)

\(=\left(2+x\right)\left(8+x\right)\)

p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)\)

24 tháng 9 2017

Bài 1 :

a ) \(6x^2+9x=3x\left(x+3\right)\)

b ) \(4x^2+8x=4x\left(x+2\right)\)

c ) \(5x^2+10x=5x\left(x+2\right)\)

d ) \(2x^2-8x=2x\left(x-4\right)\)

e ) \(5x-15y=5\left(x-3y\right)\)

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

g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)

h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)

i ) \(9x^2-18x+9=\left(3x-3\right)^2\)

k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)

l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)

m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)

n ) \(\left(x-15\right)^2=x^2-30x+15^2\)

o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)

p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)

Bài 2 :

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

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

c ) \(2x+x^2-2y-2xy=......................\)

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

e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)

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

28 tháng 9 2018

a, 3x - 3y = 3( x- y )

b, x2 - x =x(x - 1)

c, 3(x - y) - 5x(y - x)

= 3(x - y) + 5x(x - y)

= ( x - y)(3 + 5x)

d, x(y - 1) - y(y - 1)

= (x - y)(y - 1)

e, 10x(x - y)-8y( y - x)

= 10x(x - y) + 8y(x - y)

= (10y + 8x)(x - y)

f, 2x2 +5x3 +xy

= x(2x + 5x2 + y)

g, 14x2y - 21xy2 +28x2y2

= 7xy(2x - 3y + 4xy)

h, x2 - 3x + 2

= x2 - x - 2x + 2

= x(x - 1)- 2(x - 1)

= (x - 2)(x - 1)

i, x2 - x - 6

x2 + 2x - 3x - 6

x(x + 2) - 3(x + 2)

(x + 2)(x - 3)

k, x2 + 5x+6

= x2 - x + 6x + 6

=x(x - 1) + 6(x + 1)

= x(x - 1) - 6(x - 1)

= (x - 6)(x - 1)

l,x2 - 4x + 3

= x2 - x - 3x + 3

= x(x - 1) - 3(x - 1)

= (x - 3)(x - 1)

m, x2 + 5x +4

= x2 + x + 4x + 4

= x(x + 1) + 4(x + 1)

= (x + 4)(x + 1)

28 tháng 9 2018

Hướng dẫn:

a, b, c, d, e, f, g: Phương pháp phân phối đưa thừa số chung ra ngoài

h, i, k, l, m : Tách hạng tử rồi nhóm

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27 tháng 10 2021

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