Question

(2x+3)^2+(2x-3)^2+(2x+3)(4x-6)+xy

x^2+x-y^2+y

3x^2+3y^2-6xy-12

x^3-x+3x^2y+3xy^2-y+y^3

2018x^2-2019x+1=0

Answer 1

a) ( 2x +3)² + (2x-3)² + (2x+3)(4x-6) + xy

= (2x+3)² + 2(2x+3)(2x-3) + xy

= \([\) (2x+3) + (2x-3) \(]\)² + xy

= (4x)² + xy = 16x² + xy = x(16 + y)

b) x² + x - y² + y

= (x² - y² ) + ( x + y )

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

= (x+y)(x-y+1)

c) 3x² + 3y² - 6xy - 12

= 3(x² + y² - 2xy - 4)

= 3[ (x-y)² -2² ] = 3(x-y-2)(x-y+2)

d) x³ -x + 3x²y + 3xy² -y + y³

= ( x³ + 3x²y + 3xy² + y³ ) - (x + y)

= (x+y)³ - (x+y)

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

e) 2018x² - 2019x + 1 = 0

=> 2018x² - 2018x - x + 1 = 0

=> 2018x(x-1) - (x-1) = 0

=> (x-1)(2018x-1) = 0

=> \(\left[{}\begin{matrix}x-1=0\\2018x-1=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2018}\end{matrix}\right.\)