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a. \(x^2+3x+5\)
\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
=> đpcm
Giải:
a) \(x^2-6x+10\)
\(=x^2+6x+9+1\)
\(=\left(x+3\right)^2+1\)
Vì \(\left(x+3\right)^2\ge0\forall x\)
Nên \(\left(x+3\right)^2+1\ge1\forall x\)
Vậy \(\left(x+3\right)^2+1>0\forall x\).
b) \(4x-x^2-5\)
\(=-x^2+4x-4-1\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x+2\right)^2-1\)
Vì \(-\left(x-2\right)^2\le0\forall x\)
Nên \(-\left(x+2\right)^2-1\le-1\forall x\)
Vậy \(-\left(x+2\right)^2-1< 0\forall x\).
Chúc bạn học tốt!
\(\text{a) }x^2-6x+10\\ =x^2-6x+9+1\\ =\left(x^2-6x+9\right)+1\\ =\left(x^2-2\cdot x\cdot3+3^2\right)+1\\ =\left(x-3\right)^2+1\\ \text{Ta có : }\left(x-3\right)^2\ge0\forall x\\ \Rightarrow\left(x-3\right)^2+1\ge1\forall x\\ \Rightarrow\left(x-3\right)^2+1>0\forall x\left(đpcm\right)\\ \text{Vậy biểu thức luôn nhận giá trị dương }\forall x\)
\(\text{b) }4x-x^2-5\\ =-x^2+4x-4-1\\ =-\left(x^2-4x+4\right)-1\\ =-\left(x^2-2\cdot x\cdot2+2^2\right)-1\\ =-\left(x-2\right)^2-1\\ \text{Ta có : }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(x-2\right)^2\le0\forall x\\ \Rightarrow-\left(x-2\right)^2-1\le-1\forall x\\ \Rightarrow-\left(x-2\right)^2-1< 0\forall x\left(đpcm\right)\\ \text{Vậy biểu thức luôn nhận giá trị âm }\forall x\)
a) \(x^2\) − 6x + 10
= ( \(x^2\) − 6x + 9) + 1
= \(\left(x-3\right)^2\) + 1
Ta thấy : \(\left(x-3\right)^2\) \(\ge\) 0
\(\left(x-3\right)^2\) + 1 > 0 với mọi x
b) \(4x-x^2\) − 5
= − ( − 4 + \(x^2\)+ 5)
= − ( \(x^2\) − 4x + 5)
= − (\(x^2\) − 4x + 4 +1)
= − (x − 2) \(^2\) − 1
Ta thấy : − (x − 2)\(^2\) \(\le\) 0
− (x − 2)\(^2\) − < 0 với mọi x
\(x^2\)\(x^2\)\(x^2\)
a) \(x^2-6x+10\\ =x^2-6x+9+1\\ =\left(x-3\right)^2+1\)
Ta xét thấy: \(\left(x-3\right)^2\ge0\forall x\\ =>\left(x-3\right)^2+1>0\forall x\)
b) \(4x-x^2-5\\ =-\left(x^2-4x+5\right)\\ =-\left(x^2-4x+4+1\right)\\ =-\left(x-2\right)^2-1\)
Ta xét thấy:
\(-\left(x-2\right)^2\le0\forall x\\ =>-\left(x-2\right)^2-1< 0\forall x\)
a ) \(-x^2+4x-5\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1\le-1< 0\forall x\left(đpcm\right)\)
b ) \(x^4+3x^2+3=x^4+3x^2+\dfrac{9}{4}+\dfrac{3}{4}=\left(x^2+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\left(đpcm\right)\)
c ) \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3\)
Đặt \(x^2+2x+3=a\) . Khi đó , ta có :
\(x\left(x+1\right)+3=x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
\(=\left(x^2+2x+3+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
\(=\left(x^2+2x+\dfrac{7}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\left(đpcm\right)\)
a) Ta có \(2x^2-8x+13=2x^2-8x+8+5\)
\(=2\left(x^2-4x+4\right)+5\)
\(=2\left(x-2\right)^2+5\ge5\forall x\)
Giả sử trước khi làm nhé
\(a)\)\(2x^2-8x+13>0\)
\(\Leftrightarrow\)\(4x^2-16x+26>0\)
\(\Leftrightarrow\)\(\left(4x^2-16+16\right)+10>0\)
\(\Leftrightarrow\)\(\left(2x-4\right)^2+10\ge10>0\) ( luôn đúng )
Vậy ...
\(b)\)\(-2+2x-x^2< 0\)
\(\Leftrightarrow\)\(x^2-2x+2>0\)
\(\Leftrightarrow\)\(\left(x^2-2x+1\right)+1>0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2+1\ge1>0\) ( luôn đúng )
Vậy ...
Chúc bạn học tốt ~
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
x2-6x+10
=x2-6x+9+1
=(x-3)2+1>0 với mọi x (vì (x-3)2\(\ge\)0 với mọi x)
4x-x2-5
= -x2+4x-4-1
= -(x2-4x+4)-1
= -(x-2)2-1<0 với mọi x(vì -(x-2)2<0 với mọi x)
a) \(-\left(x^2-6x+10\right)=-\left(x^2-6x+9+1\right)=-\left[\left(x-3\right)^2+1\right]\le-1< 0\forall x\)
BĐT đúng
b) \(x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
BĐT đúng
c)Dấu "=" ko xảy ra???
\(=\left(4x^2+2.2x.y+y^2\right)+2\left(2x+y\right)+1+2\)
\(=\left(2x+y\right)^2+2.\left(2x+y\right).1+1+1\)
\(=\left(2x+y+1\right)^2+1\ge1>0\) (đpcm)
a. −x2 + 6x - 10
= −(x2 − 6x) − 10
= −(x2 − 2.x.3 + 32 − 9) − 10
= −(x − 3)2 + 9 − 10
= −(x − 3)2 −1
Vì (x − 3)2 ≥ 0 ∀ x ⇒ −(x − 3)2 ≤ 0 ⇒ −(x − 3)2 −1 ≤ −1
Vậy −(x − 3)2 −1 < 0 ⇒ −x2 + 6x - 10 luôn âm với mọi x
Câu a :
\(x^2+4x+5\)
\(=\left(x^2+4x+4\right)+1\)
\(=\left(x+2\right)^2+1\)
Do : \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1>0\)
Vậy \(x^2+4x+5>0\left(\forall x\right)\)
Câu b :
\(-x^4+4x^2-7\)
\(=\left(-x^4+4x^2-4\right)-3\)
\(=-\left(x^4-4x^2+4\right)-3\)
\(=-\left(x-2\right)^2-3\)
Do : \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-3\le-3< 0\)
Vậy \(-x^4+4x^2-7< 0\left(\forall x\right)\)
Wish you study well !!