\(x^2+2x+y^2-6y+4z^2-4z+11=0\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\\z=\dfrac{1}{2}\end{matrix}\right.\)

\(x^2+2x+y^2-6y+4z^2-4z+11=0\\ \Rightarrow\left(x^2+2x+1\right)+\left(y^2-6y+9\right)+\left(4z^2-4z+1\right)=0\\ \Rightarrow\left(x+1\right)^2+\left(y-3\right)^2+\left(2z-1\right)^2=0\)

Vì \(\left(x+1\right)^2\ge0;\left(y-3\right)^2\ge0;\left(2z-1\right)^2\ge0\) mà \(\left(x+1\right)^2+\left(y-3\right)^2+\left(2z-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\\\left(2z-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=-1\\y=3\\z=\dfrac{1}{2}\end{matrix}\right.\)

Bài làm:

Ta có: \(x^2+2x+y^2-6y+4z^2-4z+11=0\)

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

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

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\\\left(2z-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=3\\x=\frac{1}{2}\end{cases}}\)

Xin lỗi mk nhầm đoạn cuối là: \(\Rightarrow\hept{\begin{cases}x=-1\\y=3\\z=\frac{1}{2}\end{cases}}\) nhé:)

<=>x²+2x+1+y²-6y+9+4z²-4z+1=0

<=>(x²+2x+1)+(y²-6y+9)+(4z²-4z+1)=0

<=>(x+1)²+(y-3)²+(2z+1)²=0       (1)

Từ (1) <=> (x+1)²=0 <=> x+1=0 <=>x=-1

                  (y-3)²=0 <=>y-3=0 <=>y=3

                  (2z+1)²=0 <=>2z+1=0 <=> z=-1/2

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

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

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

VT(1) >= 0  với mọi x;y;z nên để đẳng thức (1) xảy ra thì: x = -1; y = 3; z = 1/2.

\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a,9x^2+y^2+2z^2−18x+4z−6y+20=0

⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0

⇔x=1;y=3;z=−1

b,5x^2+5y^2+8xy+2y−2x+2=0

⇔4(x+y)2+(x−1)2+(y+1)2=0

⇔x=−y;x=1y=−1⇔x=1y=−1

c,5x^2+2y^2+4xy−2x+4y+5=0

⇔(2x+y)^2+(x−1)^2+(y+2)^2=0

⇔2x=−y;x=1;y=−2

⇔x=1;y=−2

⇔(x−1)^2+(2y−3)^2+(z+2)^2=0

\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương

x² + 2x + y² - 6y + 4z² - 4z + 11 = 0

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

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

Ta có : \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\\left(2z-1\right)^2\ge0\forall z\end{cases}}\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2+\left(2z-1\right)^2\ge0\forall x,y,z\)

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

Vậy ...

6x bạn ơi

Chọn A.

Đường thẳng (Δ) đi qua M(0; 1; 2) và có một vectơ chỉ phương là 

Mặt cầu (S) có tâm I(1; 0; -2) và bán kính 

Ta có

không cắt mặt cầu (S)