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a) \(\left(3x+2y\right)\left(x^2-3xy+1\right)=3x^3-9x^2+3x+2yx^2-6xy^2+2y\)
b) (x+3)(x-3)(x+1)=(x2-9)(x+1)=x3-9x+x2-9
a) \(\left(3x+2y\right)\left(x^2-3xy+1\right)\)
\(=\left(3x+2y\right)x^2-\left(3x+2y\right)\cdot3xy+\left(3x+2y\right)\cdot1\)
\(=3x^3+2x^2y-9x^2y-6xy^2+3x+2y\)
b) \(\left(x+3\right)\left(x-3\right)\left(x+1\right)\)
\(=\left(x^2-3^2\right)\left(x+1\right)\)
\(=\left(x^2-9\right)x+\left(x^2-9\right)\cdot1\)
\(=x^3-9x+x^2-9\)
a: \(\dfrac{x^2-xy+y^2}{x^2+2xy+y^2}\cdot\dfrac{x^2+3xy+2y^2}{x^2-3xy+2y^2}\)
\(=\dfrac{x^2-xy+y^2}{\left(x+y\right)^2}\cdot\dfrac{\left(x+2y\right)\left(x+y\right)}{\left(x-2y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x^2-xy+y^2\right)\left(x+2y\right)}{\left(x-2y\right)\left(x^2-y^2\right)}\)
b: \(\dfrac{x^2+1}{3x}:\dfrac{x^2+1}{x-1}:\dfrac{x^3-1}{x^2+x}:\dfrac{x^2+2x+1}{x^2+x+1}\)
\(=\dfrac{x-1}{3x}\cdot\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{\left(x+1\right)^2}\)
\(=\dfrac{x\left(x+1\right)}{3x\left(x+1\right)^2}=\dfrac{1}{3\left(x+1\right)}\)
sau bạn đăng tách ra cho mn cùng giúp nhé
a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)
b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)
c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)
d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)
e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)
f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)
\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y
= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )
= 3x2 - 15x + 2xy - 10y - x + 5 - 2xy + 4y
= 3x2 - 16x - 6y + 5
b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 )
= [ ( 3x )2 - 22 ] - ( 8x2 + 10x + 3 )
= 9x2 - 4 - 8x2 - 10x - 3
= x2 - 10 - 7