a)Min=-3 khi x=2 và y=-3

b)Min=7/2 khi x=1/2 và y=-1

A có thể giải rõ giúp e dc k

a)\(M=x^2-2xy+2y^2-4y+2016\)

\(=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012\)

\(=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)

Dấu = khi \(\begin{cases}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y=0\\y-2=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=y\\y=2\end{cases}\)\(\Leftrightarrow x=y=2\)

Vậy Min_M=2012 khi x=y=2

b)\(N=x^2-2xy+2x+2y^2-4y+2016\)

\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-2y+1\right)+2014\)

\(=\left(x-y+1\right)^2+\left(y-1\right)^2+2014\ge2014\)

Dấu = khi \(\begin{cases}\left(x-y+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y+1=0\\y-1=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x-y+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x-1+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x=0\\y=1\end{cases}\)

Vậy Min_N=2014 khi x=0;y=1

a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)

\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)

\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)

\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)

b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)

\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)

\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)

\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)

\(\Leftrightarrow\frac{1}{3x^2-y^2}\)

b, x² +y²+z² +2x-4y-6z+14=0

<=> (x²+2x+1)+(y²-4y+4)+(z²-6z+9)=0

<=> (x+1)²+(y-2)²+(z-3)²=0

=>(x+1)²=(y-2)²=(z-3)²=0

=>x+1=y-2=z-3=0

=> x=-1; y=2; z=3

c, 2x²+y²-6x-4y+2xy+5=0

<=> (x²+y²+4+2xy-4x-4y)+(x²-2x+1)=0

<=> (x+y-2)²+(x-1)²=0

=> (x+y-2)²=(x-1)²=0

=>x+y-2=x-1=0

=>x=1; y=1

1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)

\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)

2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)

\(=(3x-y)(2xy+1)\)

3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?

\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)

\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)

4. \(x^2-9x+8=(x^2-x)-(8x-8)\)

\(=x(x-1)-8(x-1)=(x-1)(x-8)\)

5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)

6. \(x^2-6x+y-y^2\) (sai đề)

7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)

\(=x(x-y)-2(x-y)=(x-y)(x-2)\)

D = 2x² + 9y² - 6xy - 6x + 12y + 2012

= [ ( x² - 6xy + 9y² ) - 4x + 12y + 4 ] + ( x² - 2x + 1 ) + 2007

= [ ( x - 3y )² - 2( x - 3y ).2 + 2² ] + ( x - 1 )² + 2007

= ( x - 3y + 2 )² + ( x - 1 )² + 2007

\(\hept{\begin{cases}\left(x-3y+2\right)^2\\\left(x-1\right)^2\end{cases}}\ge0\forall x\Rightarrow\left(x-3y+2\right)^2+\left(x-1\right)^2+2007\ge2007\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3y+2=0\\x-1=0\end{cases}}\Rightarrow x=y=1\)

=> MinD = 2007 <=> x = y = 1

E = x² - 2xy + 4y² - 2x - 10y + 29 ( -10y mới ra đc nhé, mò mãi :v )

= [ ( x² - 2xy + y² ) - 2x + 2y + 1 ] + ( 3y² - 12y + 12 ) + 16

= [ ( x - y )² - 2( x - y ) + 1² ] + 3( y² - 4y + 4 ) + 16

= ( x - y - 1 )² + 3( y - 2 )² + 16

\(\hept{\begin{cases}\left(x-y-1\right)^2\\3\left(y-2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x-y-1\right)^2+3\left(y-2\right)^2+16\ge16\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)

=> MinE = 16 <=> x = 1 ; y = 2

F = \(\frac{3}{2x-x^2-4}\)

Để F đạt GTNN => 2x - x² - 4 đạt GTLN

Ta có : 2x - x² - 4 = -( x² - 2x + 1 ) - 3 = -( x - 1 )² - 3 ≤ -3 < 0 ∀ x

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinF = \(\frac{3}{-3}=-1\)<=> x = 1

G = \(\frac{2}{6x-5-9x^2}\)

Để G đạt GTNN => 6x - 5 - 9x² đạt GTLN

Ta có 6x - 5 - 9x² = -9( x² - 2/3x + 1/9 ) - 4 = -9( x - 1/3 )² - 4 ≤ -4 < 0 ∀ x

Đẳng thức xảy ra <=> x - 1/3 = 0 => x = 1/3

=> MinG = \(\frac{2}{-4}=-\frac{1}{2}\)<=> x = 1/3