Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x^2+y^2+2xy\right)+\left(x^2y^2+2xy+1\right)+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y\right)^2+\left(1+xy\right)^2+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y+1+xy\right)^2\) là SCP
(1+x2)(1+y2)+4xy+2(x+y)(1+xy)
= 1+y2+x2+x2y2+2xy+2xy+2(x+y)(1+xy)
=(x2+2xy+y2)+(x2y2+2xy+1)+2(x+y)(1+xy)
=(x+y)2+(xy+1)2+2(x+y)(1+xy)
=(x+y+xy+1)2
x2 + xy + y2 + 1 = (x2 + 2.x. \(\frac{y}{2}\) + (\(\frac{y}{2}\))2 ) + \(\frac{3y^2}{4}\) + 1 = (x + \(\frac{y}{2}\))2 + \(\frac{3y^2}{4}\) + 1 \(\ge\) 0 + 0 + 1 = 1> 0 với mọi x; y
Ta có:
x2+xy+y2+1=x2+xy+1/4.y2+3/4.y2+1=(x+1/2.y)2+3/4.y2+1
Mà (x+1/2.y)2 \(\ge\)0
3/4.y2>=0
1>0
Suy ra (x+1/2.y)2+3/4.y2+1>0
Hay x2+xy+y2+1>0(đpcm)
\(A=x^2-xy+y^2\)
\(\Rightarrow A=x^2-xy+\dfrac{1}{4}y^2-\dfrac{1}{4}y^2+y^2\)
\(\Rightarrow A=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\)
mà \(\left(x-\dfrac{1}{2}y\right)^2\ge0;\dfrac{3}{4}y^2\ge0\)
\(\Rightarrow A=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)
\(\Rightarrow\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\) với mọi x,y không đồng thời bằng 0
a) Ta có:
\(x^2-x+1\)
\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên
\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
\(\Rightarrow x^2-x+1>0\forall x\)
a/ \(x^2+xy+y^2+1\)=\(\left(x^2+2x\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2\right)+\dfrac{3y^2}{4}+1\)
=\(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\) \(\ge\)0
vậy....
b
\(x^2+xy+y^2+1=\left(x^2+xy+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\forall x;y\)