toán 12 nha

A = 9x² - 6xy + 5y² + 1 = (3x)² + 2.3y + y² + (2y)² + 1 = ( 3x + y)² + ( 2y )² +1 
mà ( 3x + y)² > 0 và ( 2y )² > 0 

=> ( 3x + y )² + (2y)² + 1 > 0

Vậy gtnn của A là 1 

`A=x^4-6x^3+18x^2-6xy+y^2+2012`
`=x^4-6x^3+9x^2+9x^2-6xy+y^2+2012`
`=(x^2-x)^2+(3x-y)^2+2012>=2012`
Dấu "=" xảy ra khi:
$\begin{cases}x=x^2\\y=3x\end{cases}$
`<=>` $\left[ \begin{array}{l}\begin{cases}x=0\\y=3x=0\\\end{cases}\\\begin{cases}x=1\\y=3x=3\\\end{cases}\end{array} \right.$
Vậy `min_A=2012<=>` $\left[ \begin{array}{l}x=y=0\\\begin{cases}x=1\\y=3\end{cases}\end{array} \right.$

XD moi x

\(yx^2+y=x^2+3x+5\Leftrightarrow\left(y-1\right)x^2-3x+\left(y-5\right)=0\)

dat y-1=a cho gon

\(ax^2-3x+\left(a-4\right)=0\)(1)

tim DK a de phuong trinh tren(1) co nghiem

a=0=>-3x-4=0=> x=4/3

voi a \(\ne0\)(1) phuong trinh bac 2

=>delta(x)=3^2-4a.(a-4)\(\ge0\) 

\(\Leftrightarrow9-4a^2+16a\ge0\Leftrightarrow4a^2-16a-9\le0\)

delta"(a)=4^2-4.(-9)=16+36=52=4.13

\(\orbr{\begin{cases}a_1=\frac{4-2\sqrt{13}}{4}=1-\frac{\sqrt{13}}{2}\\a_2=\frac{4+2\sqrt{13}}{4}=1+\frac{\sqrt{13}}{2}\end{cases}}\)

\(\left(1-\frac{\sqrt{13}}{2}\right)\le a\le1+\frac{\sqrt{13}}{2}\)

\(1-\frac{\sqrt{13}}{2}\le y-1\le1+\frac{\sqrt{13}}{2}\)

\(2-\frac{\sqrt{13}}{2}\le y\le2+\frac{\sqrt{13}}{2}\)