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)

Phân tích đa thức sau thành nhân tử :

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

b) \(x^3+y^3+z^3-3xyz\)

\(x^3+y^3+z^3-3xyz\)

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

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

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

Đa thức trên tương đương với đa thức:

\(\left(xy\left(x+y\right)+xyz\right)+\left(yz\left(y+z\right)+xyz\right)+\left(xz\left(x+z\right)+xyz\right)\)

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

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

xy(x + y) + yz( y + z )+ zx( z + x ) + 3xyz

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

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

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

chúc bn hok tốt

Sửa đề chút :

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

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

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

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

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

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

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

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

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

c) x³ + 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

huyen

Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)

Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

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

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

\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Hay \(A=3\cdot2x\cdot2y\cdot2z\)

\(A=24xyz\)

Biến đổi : \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3\) theo công thức tổng của hai lập pương , ta được :

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

Thay vào \(A\),ta có : \(A=\left(y^2+z^2\right).B\).Trong đó :

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

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

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

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

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

Đặt \(x-y=a,y-z=b,z-x=c\Rightarrow a+b+c=0\)

\(\Rightarrow a+b=-c\)

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

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

\(\Rightarrow a^3+b^3+3ab.\left(-c\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Vậy \(A=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

        \(x^4+4x^2+16\)

\(=\left(x^2\right)^2+2.x^2.4+4^2-4x^2\)

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

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