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\(2x^2+9y^2+3z^2+6xy-2xz+6yz\)
\(=\left(2x^2-6xy-2xz+\frac{9}{2}y^2+3yz+\frac{z^2}{2}\right)+\left(\frac{9}{2}y^2+3yz+\frac{z^2}{2}\right)+2z^2\)
\(=\left[2x^2-2x\left(3y+z\right)+\frac{9y^2+6yz+z^2}{2}\right]+\frac{9y^2+6yz+z^2}{2}+2z^2\)
\(=\left[2x^2-2.2.x.\frac{3y+z}{2}+\frac{\left(3y+z\right)^2}{2}\right]+\frac{\left(3y+z\right)^2}{2}+2z^2\)
\(=2\left[x^2-2.x.\frac{3y+z}{2}+\frac{\left(3y+z\right)^2}{4}\right]+\frac{\left(3y+z\right)^2}{2}+2z^2\)
\(=2\left(x^2-\frac{3y+z}{2}\right)^2+\frac{\left(3y+z\right)^2}{2}+2z^2\ge0\forall x;y;z\)
Ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\right]\\ b,\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\\ \Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\\ \Leftrightarrow x,y,z\in\varnothing\left[\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\right]\)
\(c,\Leftrightarrow-\left(x^2-10xy+25y^2\right)-\left(y^2-20y+100\right)-50=0\\ \Leftrightarrow-\left(x-5y\right)^2-\left(y-10\right)^2-50=0\\ \Leftrightarrow x,y\in\varnothing\left[-\left(x-5y\right)^2-\left(y-10\right)^2-50\le-50< 0\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
= x2+2x+1+y2+6y+9
= (x+1)2+(y+3)2
Vì (x+1)2 >=0 với mọi x
(y+3)2>=0 với mọi y
Do đó (x+1)2+(y+3)2>= với mọi x,y
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(=\left(x-y\right)^2+\left(x-y\right)+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-y+\frac{1}{2}\right)^2+\frac{3}{4}>0\rightarrowđpcm\)
b) \(2x^2+9y^2+3z^2+6xy+2xz+6yz\)
\(=\left(x^2+z^2+2xz\right)+6y\left(x+z\right)+9y^2+x^2+2z^2\)
\(=\left(x+z\right)^2+6y\left(x+z\right)+9y^2+x^2+2z^2\)
\(=\left(x+z+3y\right)^2+x^2+2z^2\ge0\rightarrowđpcm\)
a)