Ta có:

\(x^2+y^2+z^2-4x+2y+6z\)

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

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

Mà : \(\left(x-2\right)^2\ge0\forall x\)

        \(\left(y+1\right)^2\ge0\forall y\)

          \(\left(z+3\right)^2\ge0\forall z\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z+3\right)^2\ge0\forall x;y;z\) ( luôn đúng )

\(\Rightarrow x^2+y^2+z^2+14\ge4x-2y-6z\left(đpcm\right)\)

Dấu "=" xảy ra khi :

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

Vậy ....

Ta có (x²+4x+4)+(y²+2y+1)+(z^2+6z+9)>=0

a)x\(^2\)+10x+26+y\(^2\)+2y

=(^2+10x+25)+(y^2+2y+1)

=(x+5)^2+(y+1)^2

a. x² + 10x + 26 + y² + 2y

= x² + 10x + 25 + y² + 2y + 1

= (x + 5)² + (y + 1)² (Xem lại đề)

b. z² - 6z + 5 - t² - 4t

= z² - 6z + 9 - t² - 4t - 4

= (z - 3)² - (t² + 4t + 4)

= (z - 3)² - (t + 2)²

c. (y + 2z - 3).(y - 2z - 3) 

= (y - 3 + 2z).(y - 3 - 2z)

= (y - 3)² - (2z)²

d. (x + 2y + 3z).(2y + 3z - x)

= (2y + 3z + x).(2y + 3z - x)

= (2y + 3z)² - x²

\(x^2+y^2+z^2=4x-2y+6z-14\)

\(\Leftrightarrow x^2-4x+4+y^2+2y+1+z^2-6z+9=0\)

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

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

\(\Leftrightarrow\) \(x^2\)+    \(y^2\) +     \(z^2\) -    \(4x\)+      \(2y\) -      \(6z\) +    \(14\) \(=\) \(0\)

\(\Leftrightarrow\) (  \(x^2\) -     \(4x\) +    \(4\)  )   +      (   \(y^2\) +    \(2y\) +     \(1\) )   \(=\) \(0\)

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

\(\Leftrightarrow\) \(\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}\)

B = ( 2x + 2y - z )² + ( 2y + 2z - x )² + ( 2z + 2x - y)²

B =4x²+4y²+z²+8xy-4xz-4yz+4y²+4z²+x²+8yz-4xz-4xy+4z²+4x²+y²+8xz-4xy-4yz

B =9x²+9y²+9z²

tick cho mình nhá

a.(x+y)²-xy+1>0 với mọi y,x

b/ a. ( x + y ) ² -xy + 1 > 0 vs mọi x, y 

TK , MK ĐANG BỊ ÂM ĐIỂM

Ta có: \(x^2+y^2-4x=6z-2y-z^2-14\)

\(x^2+y^2-4x-6z+2y+z^2+14=0\)

\(\left(x^2-4x+2^2\right)+\left(y^2+2y+1\right)+\left(z^2-6z+3^2\right)=0\)

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

\(\cdot\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

\(\cdot\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\)

\(\left(z-3\right)^2=0\Rightarrow z-3=0\Rightarrow z=3\)

hok tốt!

Ta có x² + y² - 4x = 6z - 2y - z² - 14

=> x² + y² - 4x - 6z + 2y + z² + 14 = 0

=> (x² - 4x + 4) + (y² + 2y + 1) + (z² - 6z + 9) = 0

=> (x - 2)² + (y + 1)² + (z - 3)² = 0

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

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

Vậy x = 2 ; y = - 1 ; z = 3