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

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

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

X^3-3x^2+3x^2+3x-1

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

=x^3+3x-1

\(2.\left(3x-1\right).\left(2x-5\right)-\left(4x-1\right).\left(3x-2\right)\)

\(\Leftrightarrow6x-2.\left(2x-5\right)-12x^2-8x-3x+2\)

\(\Leftrightarrow12x^2-30x-4x+10-12x^2-11x+2\)

\(\Leftrightarrow-45x+12\)

2( 3x - 1 )( 2x - 5 ) - ( 4x - 1 )( 3x - 2 )

= 2( 6x² - 17x + 5 ) - ( 12x² - 11x + 2 )

= 12x² - 34x + 10 - 12x² + 11x - 2

= -23x + 8 

( 3x² - 2x + 1 )( 3x² + 2x + 1 ) - ( 3x² + 1 )²

= [ ( 3x² + 1 ) - 2x ][ ( 3x² + 1 ) + 2x ] - ( 3x² + 1 )²

= ( 3x² + 1 )² - 4x² - ( 3x² + 1 )²

= -4x²

\(M=18+4x-8y+6xy+5x^2+10y^2\)

\(=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)

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

Có \(\left(x+y\right)^2\ge0\forall xy\)

\(4\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\left(y-4\right)^2\ge0\forall y\)

\(\Rightarrow M\ge1\forall x,y\)

hay \(M>0\forall x,y\)

giải trên cymath.com í

3 ( 1 - 2x ) ( 5 - 3x )

= ( 3 - 6x ) ( 5 - 3x )

= 15 - 9x - 30x + 18x²

= 18x² - 39x + 15

3(1-2x)(5-3x)

=3(5-10x-3x+6x²)

=15-30x-9x+18x²

=15-39x+18x²

x² + y² - 3x - 3y + 2xy

= ( x² + 2xy + y² ) - ( 3x + 3y )

= ( x + y )² - 3( x + y )

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

b) ( x² - 4x )² - 2( x - 2 )² - 7 

= ( x² - 4x )² - 2( x² - 4x + 4 ) - 7 (*)

Đặt t = x² - 4x

(*) <=> t² - 2( t + 4 ) - 7

       = t² - 2t - 8 - 7

       = t² - 2t - 15

       = t² + 3t - 5t - 15

       = t( t + 3 ) - 5( t + 3 )

       = ( t + 3 )( t - 5 )

       = ( x² - 4x + 3 )( x² - 4x - 5 ) 

       = ( x² - x - 3x + 3 )( x² + x - 5x - 5 )

       = [ x( x - 1 ) - 3( x - 1 ) ][ x( x + 1 ) - 5( x + 1 ) ]

       = ( x - 1 )( x - 3 )( x + 1 )( x - 5 )

a) Ta có: \(x^2+y^2-3x-3y+2xy\)

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

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

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

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

        \(=\left(x+y-1+\sqrt{x+y+1}\right)\left(x+y-1-\sqrt{x+y+1}\right)\)

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

\(=-x^3y-3x^2-2x^2y^2-6xy+3xy+9\)

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

( x² + 2xy - 3 ) ( - xy - 3 )

= - x³y - 2x²y² + 3xy - 3x² - 6xy + 9

= - x³y - 2x²y² - 3x² - 3xy + 9