\(A=x^2+2x+9y^2-6y+2018\)

\(=x^2+2x+1+9y^2-6y+1+2016\)

\(=\left(x+1\right)^2+\left(3y-1\right)^2+2016\ge2016\forall x;y\)

Dấu ''='' xảy ra khi x = -1 ; y = 1/3 

Vậy GTNN của A bằng 2016 tại x = -1 ; y = 1/3 

(8x-3)(3x+2)-(4x+7)(x+4) = (2x+1)(5x-1)-33

(24x²-9x+16x-6)-(4x²+7x+16x+28) = (10x²+5x-2x-1)-33

24x²+7x-6-4x²-23x-28 = 10x²+3x-1-33

20x²-16x-34 = 10x²+3x-34

<=> 20x²-16x = 10x²+3x

2x²-19x=0

2x(x-19)=0

=>\(\left[{}\begin{matrix}2x=0\Rightarrow x=0\\x-19=0\Rightarrow x=19\end{matrix}\right.\)

Không chắc lắm :)

\(2x-1\left(x+2\right)-3\left(x+2\right)=0\)

\(\Rightarrow\left(2x-4\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)

(2x-1)(x+2)-3(x+2)=0

<=>2x²+3x-2-3x-6=0

<=>2x²-8=0

<=>2(x²-4)=0

<=>x²-4=0

<=>(x+2)(x-2)=0

=>\(\orbr{\begin{cases}x+2=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)

Vậy...

(2x-1)(x+2)-3(x+2)=0

(x+2)[(2x-1)-3]=0

(x+2)(2x-1-3)=0

(x+2)(2x-4)=0

TH1: x+2=0 nên x=-2

TH2: 2x-4=0 nên 2x=4 nên x=2

vậy x=-2 hoặc x=2

a) x² + xy + y - 1 = (x² - 1) + (xy + y) = (x - 1)(x + 1) + y(x + 1) = (x + 1)(x + y - 1)

b) 4 - x² + 2xy - y² = 4 - (x² - 2xy + y²) = 4 - (x - y)² = (x - y + 2)(4 - x + y) 

c) 8x² - 18y² = 2(4x² - 9y²) = 2[(2x)² - (3y)²] = 2(2x - 3y)(2x + 3y)

d) 8x³ - 4x² - 6xy - 9y² - 27y³

= (8x³ - 27y³) - (4x² + 6xy + 9y²)

= (2x - 3y)(4x² + 6xy + 9y²) - (4x² + 6xy + 9y²)

= (2x - 3y - 1)(4x² + 6xy + 9y²)

e) 4x² - x - 3 = 4x² - 4x + 3x - 3 = 4x(x - 1) + 3(x - 1) = (x - 1)(4x + 3)

f) 4x² - 8x + 3 = 4x² - 2x - 6x + 3 = 2x(2x - 1) - 3(2x - 1) = (2x - 3)(2x - 1)

cảm ơn bạn

A=\(x^2+\left(3y\right)^2+4-6xy+12y-4x+x^2+4x+4+1996\)

A=\(\left(x-3y-2\right)^2+\left(x+2\right)^2+1996\ge1996\)

Vay gtnn cua A la 1996.Dat duoc khi va chi khi \(\hept{\begin{cases}x+2=0\\x-3y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\3y=-2-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\end{cases}}}\)

Study well