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a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
Câu b:
Ta có: \(x^2 + 4y^2 + z^2 - 2x - 6z + 8y + 15\)
\(= (x^2 - 2x +1) + (4y^2 - 8y + 4) + (z^2 - 6z +9) +1\)
\(= (x-1)^2 + (2y-2)^2 + (z-3)^2 + 1\)
Mà \((x-1)^2 \geq 0; (2y-2)^2 \geq 0; (z-3)^2\geq 0\)
\(\implies\) \((x-1)^2+(2y-2)^2 +(z-3)^2\geq 0\)
\(\implies\)\((x-1)^2+(2y-2)^2 +(z-3)^2+1> 0\)
Bài làm:
Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x,y,z\right)\)
x2 + 4y2 + z2 - 2x - 6z + 8y + 15
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + ( 2y + 2 )2 + ( z - 3 )2 + 1 ≥ 1 > 0 ∀ x,y,z ( đpcm )
\(x^2+4y^2+z^2-2x-6z+8y+14=0\\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x;y;z\)
Mặt khác: \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\)
nên ta được:
\(\left\{{}\begin{matrix}x-1=0\\2y+2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=3\end{matrix}\right.\)
Vậy: ...
\(x^2+4y^2+z^2-2x-6z+8y+14=0\)
\(\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)=0\)
\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\) (1)
Do \(\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)
\(\left(1\right)\Rightarrow\) \(\left(x-1\right)^2=0;\left(2y+2\right)^2=0;\left(z-3\right)^2=0\)
*) \(\left(x-1\right)^2=0\)
\(x-1=0\)
\(x=1\)
*) \(\left(2y+2\right)^2=0\)
\(2y+2=0\)
\(2y=-2\)
\(y=-1\)
*) \(\left(z-3\right)^2=0\)
\(z-3=0\)
\(z=3\)
Vậy x = 1; y = -1; z = 3
a) \(x^2+x+2=\left(x^2+x+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\)đúng \(\forall x\in R\)
b) \(x^2-4x+10=\left(x^2-4x+4\right)+6=\left(x-2\right)^2+6\ge6>0\)đúng \(\forall x\in R\)
c) \(x\left(x-4\right)+10=x^2-4x+10\)(giải như câu b)
d) \(x\left(2-x\right)-4=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\le-3< 0\)đúng \(\forall x\in R\)
e) \(x^2-5x+2017=\left(x^2-5x+\frac{25}{4}\right)+\frac{8043}{4}=\left(x-\frac{5}{2}\right)^2+\frac{8043}{4}\ge\frac{8043}{4}>0\)đúng \(\forall x\in R\)
a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
