a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

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

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

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

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)

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

\(\Rightarrow x=1;y=-2;z=-1\)

<=>(x²-2x+1)+(3y²+12y+12)+(2z²+4z+2)=0

<=>(x-1)²+3(y+2)²+2(z+1)²=0

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

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

a. \(x^2+4y^2+z^2=2x+12y-4z-14\)

\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)

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

Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)

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

b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

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

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

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

\(\overrightarrow{n_{\left(\alpha\right)}}=\left(1;2;3\right)\)

\(\overrightarrow{n_{\left(P\right)}}=\left(2;4;6\right)\)

\(\overrightarrow{n_{\left(R\right)}}=\left(2;-4;6\right)\)

\(\overrightarrow{n_{\left(Q\right)}}=\left(1;-1;2\right)\)

\(\overrightarrow{n_{\left(S\right)}}=\left(1;-1;2\right)\)

Tích vô hướng của \(\overrightarrow{n_{\left(\alpha\right)}}\) với cả 4 vecto kia đều khác 0 nên ko mặt phẳng nào vuông góc với \(\left(\alpha\right)\)

Bạn coi lại đề bài

Ta có:

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

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

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

Vì \(\left(z-1\right)^2\ge0\) với mọi z

\(\left(y-2\right)^2\ge0\) với mọi y

\(\left(x-z+2\right)^2\ge0\) với mọi x, z

Suy ra \(\left(z-1\right)^2+\left(y-2\right)^2+\left(x-z+2\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left(z-1\right)^2=0\\\left(y-2\right)^2=0\\\left(x-z+2\right)^2=0\end{matrix}\right.\)

Hay \(\left(z-1\right)^2+\left(y-2\right)^2+\left(x-z+2\right)^2=0\) khi \(\left[{}\begin{matrix}\left(z-1\right)^2=0\\\left(y-2\right)^2=0\\\left(x-z+2\right)^2=0\end{matrix}\right.\)

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

Vậy \(x=-1\); \(y=2\); \(z=1\)

a)\(\left|2x-3y\right|+\left|2y-4z\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\forall x;y\\\left|2y-4z\right|\ge0\forall y;z\end{matrix}\right.\) \(\Rightarrow\left|2x-3y\right|+\left|2y-4z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|2y-4z\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\2y=4z\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=\dfrac{y}{4}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x+y+z}{6+4+2}=\dfrac{7}{12}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{12}.6=\dfrac{7}{2}\\y=\dfrac{7}{12}.4=\dfrac{7}{3}\\z=\dfrac{7}{12}.2=\dfrac{7}{6}\end{matrix}\right.\)

b)\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|x-3\right|\ge0\\\left|x-4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|x-3\right|=0\\\left|x-4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)

Vì \(2\ne3\ne4\) nên \(x\in\varnothing\)

c)

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+8\right|+\left|x+9\right|\)

Với mọi \(x\ge0\) ta có:

\(\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\\left|x+8\right|=x+8\\\left|x+9\right|=x+9\end{matrix}\right.\)\(\Leftrightarrow x+1+x+2+...+x+8+x+9=x-1\)

\(\Leftrightarrow9x+90=x-1\)

\(\Leftrightarrow9x=x-89\)

\(\Leftrightarrow-8x=89\)

\(\Leftrightarrow x=\dfrac{89}{-8}\left(KTM\right)\)

Với mọi \(x< 0\) ta có:

\(\left\{{}\begin{matrix}x+1=-x-1\\x+2=-x-2\\x+8=-x-8\\x+9=-x-9\end{matrix}\right.\) \(\Leftrightarrow\left(-x-1\right)+\left(-x-2\right)+...+\left(-x-8\right)+\left(-x-9\right)=x-1\)

\(\Leftrightarrow-9x-90=x-1\)

\(\Leftrightarrow-9x=x+89\)

\(\Leftrightarrow-10x=89\)

\(\Leftrightarrow x=\dfrac{89}{-10}\left(TM\right)\)

d)\(\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\\ \left|5y-2z\right|\ge0\\ \left|2z-6\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|5y-2z\right|=0\\\left|2z-6\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=3\\y=\dfrac{6}{5}\\x=\dfrac{9}{5}\end{matrix}\right.\)