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a) \(A=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) với mọi x
b) \(B=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\) với mọi x
c) \(x^2+xy+y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\) với mọi x,y
d) bạn kiểm tra lại đề câu d) nhé:
\(x^2+4y^2+z^2-2x-6y+8z+15\)
\(=\left(x-1\right)^2+\left(2y-\frac{6}{4}\right)^2+\left(z+4\right)^2-\frac{13}{4}\)
a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
c) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)
\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2+6\right]\)
\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2\right]-6\le-6< 0\forall x\)
Bài làm:
a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)
\(=-\left(2x+1\right)^2-1\le-1< 0\left(\forall x\right)\)
=> đpcm
b) \(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2-8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x\right)\)
=> đpcm
a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)
\(=-\left(2x+1\right)^2-1\)
Vì \(-\left(2x+1\right)^2\le0\forall x\)\(\Rightarrow\)\(-\left(2x+1\right)^2-1\le-1\forall x\)
\(\Rightarrow\)\(-\left(2x+1\right)^2-1< 0\forall x\)
\(\Rightarrow\)\(-4x^2-4x-2< 0\forall x\)( ĐPCM )
b) Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\)\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)
\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x,y,z\)
\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\forall x,y,z\)( ĐPCM )
Có : x^2+y^2+z^2+4x-2y-4z+10
= (x^2+4x+4)+(y^2-2y+1)+(z^2-4x+4)+1
= (x+2)^2+(y-1)^2+(z-2)^2+1 >= 1
=> (x+2)^2+(y-1)^2+(z-2)^2 luôn dương với mọi x,y,z
\(x^2+y^2+z^2+4x-2y-4z+10\)
\(=\left(x^2+4x+4\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)+1\)
\(=\left(x+2\right)^2+\left(y-1\right)^2+\left(z-2\right)^2+1\)
Vì \(\hept{\begin{cases}\left(x+2\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\left(x+2\right)^2+\left(y-1\right)^2+\left(z-2\right)^2\ge0\)
\(\Rightarrow\)\(\left(x+2\right)^2+\left(y-1\right)^2+\left(z-2\right)^2+1>0\)
\(\Rightarrow\)\(đpcm\)
A) x2+4y22+z22-4x-6z+15>0 <=> (x2-2×2×x+22)+4y2+(z2-2×3×z+32) +(15 -22-32) >0
<=>(x-2)2+4y22+(z-3)2
B) giải
(2X)2+ 2×2X×1 +1 >=0 với mọi X ( (2x+1)2 )
=> (2x+1)2+2 >0
Ta có : x2 + 2x + 2
= x2 + 2x + 1 + 1
= (x + 1)2 + 1 \(\ge1\forall x\)
Vậy x2 + 2x + 2 \(>0\forall x\)
Ta có : x2 + 2x + 2
=> x2 + 2x + 1 + 1
=> ( x + 1)2 + 1 > 1\(\forall x\)
Vậy x2 + 2x + 2 > \(0\forall x\)
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
\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
\(\Rightarrow x^2-x+1>0\forall x\)