I don't nơ

đa thức và nhân tử là gì

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

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

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

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

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

Chúc bạn học tốt.

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

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

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

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

(x+y)³- 1 - 3xy(x+y-1)

=(x+y)³  -1³ -3xy(x+y-1)

=(x+y-1)³  -3xy(x+y-1)

=(x+y-1)  [(x+y-1)-3xy]

=(x+y-1) [x+y-1-3xy]

chúc bạn học tốt nha!

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

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

\(=x^3+3xy\)

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

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

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

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

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

mik nghĩ tới đây đủ r ạ, nếu đổi x^3+y^3 thì nó sẽ lại ra ngược (x+y)^3 -3xy(x+y). Sai thì ibb ạ.Cảm ơn

a) 4(x²-y²)-8(x-ay)-4(a²-1)

    => 4x²-4y²-8x+8ay-4a²+4

    => 4(x²-y²-2x+2ay-a²+1)

c) a⁵+a⁴+a³ +a² +a+1

    => a(a⁴+a³+a²+a+1)+1

a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)

\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)

\(=\left(x-y\right)\left(a+b+c\right)\)

b: \(a^m-a^{m+2}\)

\(=a^m-a^m\cdot a^2\)

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

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

\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)

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

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

\(\left(x-1\right)^2-\left(x+1\right)^2\\ =\left(x-1-x-1\right)\left(x-1+x+1\right)\\ =-2.2x=-4x\)

a: =(2x)^3+3*(2x)^2*3y+3*2x*(3y)^2+(3y)^3

=(2x+3y)^3

b: (x-y)^3-(x+y)^3

=(x-y-x-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]

=-2y*[x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2]

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

c: (x+1)^3+(x-1)^3

=(x+1+x-1)[(x+1)^2-(x+1)(x-1)+(x-1)^2]

=2x*[x^2+2x+1-x^2+1+x^2-2x+1]

=2x(x^2+3)

d: =(x-1-x-1)(x-1+x+1)

=2x*(-2)=-4x

\(8x^3+36x^2y+54xy^2+27y^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)

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

\(------\)

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

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

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

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

\(------\)

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

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

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

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

\(------\)

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

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

\(=-2.2x=-4x\)

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

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

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

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

giúp mình câu khác được ko? câu này mình biết làm òi

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