a, TK:

(x lẻ do \(2y^2-8y+3=2\left(y^2-4y\right)+3=x^2\) lẻ)

\(b,\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+4y+4\right)=9\\ \Leftrightarrow\left(x-2\right)^2+\left(y+2\right)^2=9\)

Vậy pt vô nghiệm do 9 ko phải tổng 2 số chính phương

câu A thiếu đề

B=\(x^2-2x+2017=\left(x-1\right)^2+2016>=2016\)

Min B=2016 khi x-1=0<=>x=1

+)D=\(-2x^2+4x+2017=-2\left(x^2-2x+1\right)+2019=-2\left(x-1\right)^2+2019< =2019\)

=>Max D=2019, dấu '=' xảy ra khi x-1=0<=>x=1

Bổ sung câu A. \(A=x^2+2xy+3y^2-4y+2017\)

\(A=x^2+2xy+3y^2-4y+2017\)

\(A=\left(x^2+2xy+y^2\right)+\left(2y^2-4y+2\right)+2015\)

\(A=\left(x^2+2xy+y^2\right)+2\left(y^2-2y+1\right)+2015\)

\(A=\left(x+y\right)^2+2\left(y-1\right)^2+2015\ge2015\)

Vậy A_min=2015 <=> x=-1 và y=1

a) x²- 2x - 4y² - 4y = (x² - 2x + 1) - (4y² + 4y + 1) = (x - 1)² - (2y + 1)² = (x - 1 - 2y - 1)(x - 1 + 2y + 1) = (x - 2y - 2)(x + 2y)

b) x³ - 4x² + 12x - 27 = (x³ - 3x²) - (x² - 3x) + (9x - 27) = x²(x - 3) - x(x - 3) + 9(x - 3) = (x² - x + 9)(x - 3)

d) x⁴ - 2x³ + 2x - 1 = (x⁴  - 2x³ + x²) - (x² - 2x + 1) = (x² - x)² - (x - 1)² = (x² - x - x + 1)(x² - x + x - 1)

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

e) x⁴ + 2x³ - 4x - 4 = (x⁴ + 2x⁴ + x²) - (x² + 4x + 4) = (x² + x)² - (x + 2)² = (x² + x - x - 2)(x²  + x + x + 2) = (x² - 2)(x² + 2x + 2)

d)  D = x⁴ - 6x² + 10

D = (X²)² - 2. x². 3 + 3² + 1

D = (x² - 3)² + 1

(x² - 3)²  >= 0 với mọi x

(x² - 3)² + 1 >=1 với moi5 x

Vậy GTNN của D là 1

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

\(=x^2+x+2x+2\)

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

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

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

    a, x^2 - y^2 - 2x - 2y

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

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

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

     b, ( 6x + 3 ) - ( 2x - 5 ) ( 2x + 1 )

= 

     c, x^2 - 2x - 4y^2 - 4y

= (x^2 - 4y^2) + (-2x - 4y)

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

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