A = x²+2y² 

    = (x²+2x+1) + 2(y²+2y+1) -2x-4y-3

   = (x+1)² + 2(y+1)² - 2(x+2y) -3

   = (x+1)² + 2(y+1)² -5

 _{\(\Rightarrow A_{min}=\)}\(-5\Leftrightarrow\orbr{\begin{cases}x=-1\\y=-1\end{cases}}\) 

Ta có:

P = \(3x^2+5y^2-4\left(4x+y+xy\right)+31\)

   = \(3x^2+5y^2-4x^2-4y-4xy+31\)

   = \(5y^2-x^2-4y-4xy+31\)

   = \(\left(4y^2-4xy+x^2\right)+\left(y^2-4y+4\right)-2x^2+27\)

   = \(\left(2y-x\right)^2+\left(y-2\right)^2-2x^2+27\)

Do  \(\left(2y-x\right)^2+\left(y-2\right)^2-2x^2+27\) \(\ge\)27 nên

Min P = 27.

Dấu = xảy ra khi \(\hept{\begin{cases}2y-x=0\\y-2=0\\x=0\end{cases}}\)=> (x,y) = (4,2) ; (0,0)

pt <=> 8x+12-7x+7=40x-8+166

<=> x+19=40x+158

<=> 39x=-139

<=> x=-139/39

=> 3x²-2x+6x-4=0

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

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

=>x=-2 hoặc x=2/3

x = -2 , x= -2/3

a) x² - 7x + 10 = 0

x² - 5x - 2x + 10 = 0

(x² - 5x) - (2x - 10) = 0

x (x - 5) - 2 (x - 5) = 0

(x - 5) (x - 2) = 0

\(\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}}\)

\(\orbr{\begin{cases}x=0+5\\x=0+2\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=2\end{cases}}\)

Vậy x = 5 hoặc x = 2

b) tương tự

a) (x - 5)(x-2) = 0

=> x=2 , x=5

b) ( x+2)(7x-1) =0

=> x= -2 , x= 1/7