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

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

\(3x=2y=z\Rightarrow\frac{z}{6}=\frac{x}{2}=\frac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{z}{6}=\frac{x}{2}=\frac{y}{3}=\frac{x+y+z}{6+2+3}=\frac{99}{11}=9\)

\(\Rightarrow\hept{\begin{cases}z=54\\x=18\\y=27\end{cases}}\)

\(\frac{2x}{1}=\frac{-3y}{-1}=\frac{4z}{-2}\)

áp dụng tính chất dãy tỉ số bằng nhau  ta có

\(\frac{2x}{1}=\frac{-3y}{-1}=\frac{4z}{-2}=\frac{2x-3y+4z}{1+-1-2}=\frac{48}{-2}=-24\)

\(\Rightarrow\hept{\begin{cases}x=-12\\y=-8\\z=-12\end{cases}}\)

\(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}=\frac{x+y+z-6}{12}=\frac{24}{12}=2\)

\(\Leftrightarrow\hept{\begin{cases}x=7\\y=10\\z=13\end{cases}}\)

+) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

=> \(\hept{\begin{cases}\frac{x^2}{9}=4\\\frac{y^2}{16}=4\end{cases}}\) => \(\hept{\begin{cases}x^2=4.9=36\\y^2=4.16=64\end{cases}}\) => \(\hept{\begin{cases}x=\pm6\\y=\pm8\end{cases}}\)

Vậy ...

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

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

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

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

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

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

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

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

1. Áp dụng TCDTSBN ta có:

$\frac{x-1}{3}=\frac{y-2}{4}=\frac{z+5}{6}=\frac{x-1+(y-2)-(z+5)}{3+4-6}$

$=\frac{x+y-z-8}{1}=\frac{8-8}{1}=0$

$\Rightarrow x-1=y-2=z+5=0$

$\Rightarrow x=1; y=2; z=-5$

2.

Có:

$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}=\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}$

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

$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}=\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}=\frac{2x+2+3y+9+4z+20}{4+12+24}=\frac{2x+3y+4z+31}{40}=\frac{9+31}{40}=1$

Suy ra:

$x+1=2.1=2\Rightarrow x=1$

$y+3=1.4=4\Rightarrow y=1$

$z+5=6.1=6\Rightarrow z=1$

$

3.

Có:

$\frac{x+1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}=\frac{3x+3}{9}=\frac{2y+4}{-8}=\frac{4z-12}{20}$

Áp dụng TCDTSBN:

$\frac{x+1}{3}=\frac{y+2}{-4}=\frac{z-3}{5}=\frac{3x+3}{9}=\frac{2y+4}{-8}=\frac{4z-12}{20}=\frac{3x+3+2y+4+4z-12}{9+(-8)+20}=\frac{3x+2y+4z-5}{21}=\frac{47-5}{21}=2$

Suy ra:

$x+1=3.2=6\Rightarrow x=5$

$y+2=(-4).2=-8\Rightarrow y=-10$

$z-3=5.2=10\Rightarrow z=13$

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