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

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

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 

bó tay dù sao mk cũng muốn bạn tick cho mk nha

A=-(x²+8x+16)+21<=21  (tự làm tiếp)

B=-(x²-2x+1)-(4y²+4y+1)+7

=-(x-1)²-(2y+1)²+7<=7

\(A=5-8x-x^2\)

\(A=-x^2-8x+5\)

\(-A=x^2+8x-5\)

\(-A=x^2+4x+4x+16-21\)

\(-A=x.\left(x+4\right)+4.\left(x+4\right)-21\)

\(-A=\left(x+4\right).\left(x+4\right)-21\)

\(A=-\left(x+4\right)^2-21\le-21\)

Dấu = xảy ra khi A = -21 \(\Leftrightarrow-\left(x+4\right)^2-21=-21\)

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

Vì x²>_0=>x²+1>0

              =>x²+5>0

=>(x²+1).(x²+5)>0

=>0>0

=>Vô lí

Vậy không có giá tri của x thoả mãn đề bài.