Ta có : 9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0 

<=> 9x² - 18x + 9 + y² - 6y + 9 + 2z² + 4z + 2 = 0 

<=> 9(x² - 2x + 1) + (y² - 6y + 9) + 2(z² + 2z + 1) = 0 

<=> 9(x - 1)² + (y - 3)² + 2(z + 1)² = 0 (*)

Vì \(9\left(x-1\right)^2\ge0\forall x\in R\)

    \(\left(y-3\right)^2\ge0\forall y\in R\)

     \(2\left(z+1\right)^2\ge0\forall z\in R\)

Nên : pt (*) <=> \(\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)

Vậy pt có nhiệm (x;y;z) = (1;3;-1)

k mik , mik chỉ cko 

9x^2+y^2+2z^2-18x+4z-6y+20=0 <=>(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)=0 <=>9(x^2-2x+1)+(y-3)^2+2(z^2+2z+1)=0 <=>9(x-1)^2+(y-3)^2+2(z+1)^2=0 <=>x=1;y=3;z=-1

9x^2+ y^2 + 2z^2 - 18x + 4z - 6y + 20 = 0

<=>9x²-18x+9+y²-6y+9+2z²+4z+2=0

<=>(3x-3)²+(y-3)²+2.(z²+2z+1)=0

<=>(3x-3)²+(y-3)²+2.(z+1)²=0

<=>3x-3=0 và y-3=0 và z+1=0

<=>x=1 và y=3 và z=-1

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=1\\y=3\\z=-1\end{cases}\)

Bài 1a/

\(\frac{1}{1+x+xy}=\frac{xyz}{xyz+x+xy}=\frac{yz}{1+y+yz}\)

\(\frac{1}{1+z+xz}=\frac{y}{y+yz+xyz}=\frac{y}{1+y+yz}\)

Vậy \(M=\frac{1}{1+y+yz}+\frac{y}{1+y+yz}+\frac{yz}{1+y+yz}=1\)

Chiều về làm tiếp

Bài 1b:Lời giải này chủ yếu nhờ dự đoán trước Min là 2011/2012 đạt được khi x=2012

Ta có \(P=\frac{2012x^2-2.2012x+2012^2}{2012x^2}=\frac{\left(x-2012\right)^2+2011x^2}{2012x^2}\ge\frac{2011x^2}{2012x^2}=\frac{2011}{2012}\)

Bài 2: Dùng phân tích thành bình phương

\(10x^2+y^2+4z^2+6x-4y-4xz+5=\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)\)

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

\(\Rightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}}\)

Bài 3:

a/\(pt\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\Leftrightarrow x=-6,x=5\)

b/ta phân tích vế trái thành:\(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)

ta co 9(x^2-2x+1) +( y^2 -6y +9) + 2(z^2 + 2z +1) = 0

suy ra 9(x-1)^2 + (y - 3 )^2 + 3( z-1)^2 = 0

suy ra x-1=0 ; y-3 =0 ; z-1=0

suy ra x=1;y=3; z=1

9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

<=> 9x² - 18x + 9 + y² - 6y + 9 + 2x² + 4z + 2 = 0

<=> 9(x² - 2x + 1) + (y - 3)² + 2(z² + 2z + 1) = 0

<=> 9(x - 1)² + (y - 3)² + 2(z + 1)² = 0

<=> \(\left\{\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)

<=> \(\left\{\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)(*)

Vì \(\left(x-1\right)\ge0;\left(y-3\right)^2\ge0;\left(z+1\right)^2\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\y=3\\z=-1\end{cases}}}\)

pt ⇔ ( 9x² - 18x + 9 ) + ( y² - 6y + 9 ) + ( 2z² + 4z + 2 ) = 0

    ⇔ 9( x² - 2x + 1 ) + ( y - 3 )² + 2( z² + 2z + 1 ) = 0

    ⇔ 9( x - 1 )² + ( y - 3 )² + 2( z + 1 )² = 0

Vì \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\2\left(z+1\right)^2\ge0\forall z\end{cases}}\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)

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

Vậy