K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

2 tháng 8 2016

a)(x+y)2-(x-y)2

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

=2y.2x=4xy

b)(3x+1)2-(x+1)2

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

=2x.(4x+2)

=4x(2x+1)

c) x3+y3+z3-3xyz

= (x+y)3- 3xy(x+y) +z3-3xyz

=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)

=(x+y+z)(x2+y2+z2-xy-xz-yz)

4 tháng 8 2016

Phân tích đa thức sau thành nhân tử :

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

b) \(x^3+y^3+z^3-3xyz\)

30 tháng 5 2017

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

\(\Leftrightarrow\left[\left(x+y\right)+\left(x-y\right)\right]\left[\left(x+y\right)-\left(x-y\right)\right]\)

\(\Leftrightarrow\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(\Leftrightarrow2x.2y=4xy\)

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

\(\Leftrightarrow\left[\left(3x+1\right)+\left(x+1\right)\right]\left[\left(3x+1\right)-\left(x+1\right)\right]\)

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

\(\Leftrightarrow\left(4x+2\right).2x\)

\(\Leftrightarrow8x^2+4x\)

\(\Leftrightarrow x\left(8x+4\right)\)

18 tháng 12 2018

nếu làm đến đoạn (4x + 2). 2x đó rồi dừng cx đc phải ko

9 tháng 10 2018

Sửa đề chút :

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

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

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

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

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

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

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

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

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

9 tháng 10 2018

c) x+ y3 + z3 - 3xyz

= x3 + 3x2y + 3xy2 + y3 + z3 - 3xyz - 3x2y - 3xy2

= (x+y)3 + z3  - 3xy.( z+x+y)

= (x+y+z).[(x+y)2 - (x+y).z + z2 ] - 3xy.(x+y+z)

= (x+y+z). ( x2 + 2xy + y2 - xz - yz + z2 - 3xy)

= (x+y+z) .(x2 + y2 + z2 - xy - xz -yz)

e) (a+b-c)2 - (a-c)2 - 2ab + 2bc

= (a+b-c - a+c).(a+b+c+a-c) - 2b.(a-c)

= b.(2a+b) - 2b.(a-c)

= b.(2a+b - a +c)

= b.( a+b+c)

xl bn nha! mk chỉ nghĩ đk 2 câu thoy, 1 câu bn kia làm r! 2 câu còn lại bn đợi người tiếp theo làm nhé

2 tháng 7 2021

a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz

= xy(X + y + z)  + yz(x + y + z) + xz(X + y + z)

= (x + y +z)(xy + yz+ xz)

b) xy(x + y) - yz(y + z) - xz(z - x)

= x2y + xy2 - y2z - yz2 - xz2 + x2z

= x2(y + z) - yz(y + z) + x(y2 - z2)

= x2(y + z) - yz(y + z) + x(y + z)(y - z)

= (y + z)(x2 - yz + xy - xz)

= (y + z)[x(x + y) - z(x + y)]

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

c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)

 = x(y - z)(y + z) + yz2 - yx2 + x2z - y2z

= x(y - z)(y + z) - yz(y - z) - x2(y - z)

= (y - z)((xy + xz - yz - x2)

= (y - z)[x(y - x) - z(y - x)]

= (y - z)(x - z)(y -x) 

28 tháng 8 2018

a) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)

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

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

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

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

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

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

   = 3(y2+z2)(x4+x2y2-x2z2-y2z2)

   = 3(y2+z2)[x2(x2+y2)-z2(x2+y2)]

   = 3(y2+z2)(x2-z2)(x2+y2)

   = 3(y2+z2)(x-z)(x+z)(x2+y2)

b) \(\left(x+y\right)^3-x^3-y^3\)

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

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

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

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

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

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

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

\(=\left(x+y\right)\left(x^2+2xy+y^2+3xz+3yz+3z^2-x^2+xy-y^2\right)\)

  = (x+y)[3xy+3xz+3yz+3z

  = 3(x+y)(xy+xz+yz+z2)

  = 3(x+y)[x(y+z)+z(y+z)]

  = 3(x+y)(x+z)(y+z)

28 tháng 8 2018

a) \(\left(x^2+y^2\right)^3+\left(z^2-x^3\right)-\left(y^2+z^2\right)^3\)

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

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

b) \(\left(x+y\right)^3-x^3-y^3\)

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

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

\(=3x^2y+3xy^2\)

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

\(=x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^2-x^3-y^3-z^3\)

\(=3x^2y+3x^2z+3xy^2+3xy^2+6xyz+3xz^2+3y^2z+3yz^2\)

P/s: Ko chắc

23 tháng 9 2016

a) x3 + (a+b+c)x2+ (ab+ac+bc)x +abc

= x3 +ax2+bx2+cx2+abx+acx+bcx+abc

=x3+cx2+abx+abc+ax2+acx+bx2+bcx

=x2 (x+c) + ab (x+c) +ax (x+c) +bx (x+c)

= (x+c) (x2+ab+ax+bx)

= (x+c) { x(x+b)+a(x+b)}

=(x+c) (x+b) (x+a)

4 tháng 8 2016

d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

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

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

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

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

e)Đặt \(x^2+3x=a\)

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

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

\(=a^2-3a+a-3-5\)

\(=a^2-2a-8\)

\(=a^2+2x-4x-8\)

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

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

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

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

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

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

4 tháng 8 2016

\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)

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

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