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\(3x^2+3y^2+6x-12y+15=\left(3x^2+6x+3\right)+\left(3y^2-12y+12\right)\)
\(=3.\left(x^2+2x+1\right)+3.\left(y^2-4y+4\right)\)
\(=3.\left(x+1\right)^2+3.\left(y-2\right)^2\)
\(=3.\left(\left(x+1\right)^2+\left(y-2\right)^2\right)\)
\(\Rightarrow3.\left(\left(x+1\right)^2+\left(y-2\right)^2\right)=0\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)
Mà \(\left(x+1\right)^2\ge0,\forall x\inℝ\)
\(\left(y-2\right)^2\ge0,\forall y\inℝ\)
\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
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\(3x^2+6x+3+3y^2-12y+12=0\)
\(3\left(x^2+2x+1\right)+3\left(y^2-4y+4\right)=0\)
\(3\left(x+1\right)^2+3\left(y-2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
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b: \(x^2-6x+xy-6y\)
\(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x-6\right)\left(x+y\right)\)
c: \(2x^2+2xy-x-y\)
\(=2x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(2x-1\right)\)
e: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)
\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)
\(\Rightarrow x=1;y=-2;z=-1\)
<=>(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 <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(H=2x^2+9y^2-6xy-6y-12y+2004\)
\(\Rightarrow2H=4x^2+18y^2-12xy-12x-24y+4008\)
\(=\left(4x^2-12xy+9y^2\right)+9y^2-12x-24y+4008\)
\(=\left(2x-3y\right)^2-6\left(2x-3y\right)+9+9y^2-42y+49+3950\)
\(=\left(2x-3y-3\right)^2+\left(3y-7\right)^2+3950\ge3950\)
\(\Rightarrow2H\ge3950\)
\(\Rightarrow H\ge1975\)
Dấu "=" tại \(\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)
\(J=x^2+xy+y^2-3x-3y+1999\)
\(=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}-3x-3y+1999\)
\(=\left(x+\frac{y}{2}\right)^2-3\left(x+\frac{y}{2}\right)+\frac{9}{4}+3\left(\frac{y^2}{4}-\frac{y}{2}+\frac{1}{4}\right)+1996\)
\(=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y}{2}-\frac{1}{2}\right)^2+1996\ge1996\)
Dấu "=" tại \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)
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TL
x=3/4-3*căn bậc hai(5)/4, x=3*căn bậc hai(5)/4+3/4
-y^2+8*y-x^2+2*x+15
-y^2+7*y-x^2+3*x+21
-y^2+(2*x+12)*y-x^2-4*x+45
HT
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13: =x(a^3-1)-b(a-1)
=x(a-1)(a^2+a+1)-b(a-1)
=(a-1)(a^2x+a*x+x-b)
12: =(x-y)(x+y)-(x-y)
=(x-y)(x+y-1)
10: =3(x^2-4y^2)
=3(x-2y)*(x+2y)
7: =x^2-x-5x+5=(x-1)(x-5)
8: =x^2+3x+4x+12=(x+3)(x+4)
9: =2x^2-6x-x+3=(x-3)(2x-1)
\(3x^2+3y^2+6x-12y+15=0\)
\(\Rightarrow3.\left(x^2+y^2+2x-4y+5\right)=0\Rightarrow x^2+y^2+2x-4y+5=0\)
\(\Rightarrow x^2+y^2+2x-4y+1+4=0\)
\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x+1\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2\ge0\)
Mà \(\left(x+1\right)^2+\left(y-2\right)^2=0\)nên để thỏa mãn đẳng thức thì
\(\left(x+1\right)^2=\left(y-2\right)^2=0\) <=> x=-1 và y=2