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

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

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

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

=>x-3=0

hay x=3

\(\left(x-3\right)\left(2x^2+3\right)=0\\ \Rightarrow x-3=0\left(vì.2x^2+3>0\right)\\ \Rightarrow x=3\)

( x − 3 ) ( 2 x 2 + 3) = 0

⇒ x − 3 = 0 ( vì . 2 x 2 + 3 > 0 )

⇒ x = 3

\(B=-4x^2+12x-11\\ =-\left(\left(2x\right)^2-12x+11\right)\\ =-\left(\left(2x\right)^2-2.2x.3+9+2\right)\\ =-\left(2x-3\right)^2-2< 0\)

(vì \(\left(2x-3\right)^2\ge0\forall x\Rightarrow-\left(2x-3\right)^2\le0\forall x\Rightarrow-\left(2x-3\right)^2-2< 0\))

C=-2x^2+2x-5

=-2(x^2-x+5/2)

=-2(x^2-x+1/4+9/4)

=-2(x-1/2)^2-9/2<=-9/2<0 với mọi x

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

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

\(x^2-4x+3=0\\ \Rightarrow\left(x^2-3x\right)-\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)-\left(x-3\right)=0\\ \Rightarrow\left(x-1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(A=x^3-2x+n\)

\(B=n-2\)

\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)

Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)

Để \(A\text{⋮}B\)

⇒ \(n+4=0\)

⇒ \(n=-4\)

\(\Leftrightarrow x\in\left\{0;43\right\}\)

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

\(\Leftrightarrow2x^2-2x-x^2-2x+4=0\)

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

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)