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a) \(x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1>0\forall x\)
b) \(4x-x^2-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\forall x\)
Chứng minh rằng:
a, x^2-4x>-5 với mọi số thực x
b, Chứng minh 2x^2+4y^2-4x-4xy+5>0 với mọi số thực x;y
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Xét \(x^2-4x+4=\left(x-2\right)^2\ge0\)
<=> \(x^2-4x\ge-4>-5\)
b) \(2x^2+4y^2-4x-4xy+5\)
= \(\left(x^2-4x+4\right)+\left(x^2-4xy+4y^2\right)+1\)
= \(\left(x-2\right)^2+\left(x-2y\right)^2+1\ge1>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)
2) \(B=x^2+6x+11=\left(x+3\right)^2+2\ge2>0\left(\forall x\right)\)
3) \(C=4x^2+4x-2=\left(2x+1\right)^2-2\ge-2\) chưa chắc nhỏ hơn 0
4) \(D=-x^2-6x-11=-\left(x+3\right)^2-2\le-2< 0\left(\forall x\right)\)
5) \(E=-4x^2+4x-2=-\left(2x-1\right)^2-1\le-1< 0\left(\forall x\right)\)
1. \(A=x^2+2x+2=\left(x+1\right)^2+1\)
Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(x+1\right)^2+1\ge1\)
=> Đpcm
2. \(B=x^2+6x+11=\left(x+3\right)^2+2\)
Vì \(\left(x+3\right)^2\ge0\forall x\)\(\Rightarrow\left(x+3\right)^2+2\ge2\)
=> Đpcm
3. \(C=4x^2+4x-2=-\left(4x^2-4x+2\right)\)
\(=-\left(4\left(x-\frac{1}{2}\right)^2+1\right)\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow4\left(x-\frac{1}{2}\right)^2+1\ge1\)
\(\Rightarrow-\left(4\left(x-\frac{1}{2}\right)^2+1\right)\le1\)
=> Đpcm
4,5 làm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(-x^2+4x-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-4x+4+1\right)\)
\(=-\left(x-2\right)^2-1< 0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(x^4\ge0\forall x\)
\(3x^2\ge0\forall x\)
Do đó: \(x^4+3x^2\ge0\forall x\)
\(\Leftrightarrow x^4+3x^2+3>0\forall x\)
c: Ta có: \(\left(x^2+2x+3\right)=\left(x+1\right)^2+2>0\forall x\)
\(x^2+2x+4=\left(x+1\right)^2+3>0\forall x\)
Do đó: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a,A=2x^2+2x+1=\left(x^2+2x+1\right)+x^2=\left(x+1\right)^2+x^2\\ Mà:\left(x+1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x+1\right)^2+x^2>0\forall x\in R\\ Vậy:A>0\forall x\in R\)
2:
a: =-(x^2-3x+1)
=-(x^2-3x+9/4-5/4)
=-(x-3/2)^2+5/4 chưa chắc <0 đâu bạn
b: =-2(x^2+3/2x+3/2)
=-2(x^2+2*x*3/4+9/16+15/16)
=-2(x+3/4)^2-15/8<0 với mọi x
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)
b: \(4y^2+2y+1\)
\(=4\left(y^2+\dfrac{1}{2}y+\dfrac{1}{4}\right)\)
\(=4\left(y^2+2\cdot y\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{3}{16}\right)\)
\(=4\left(y+\dfrac{1}{4}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall y\)
c: \(-2x^2+6x-10\)
\(=-2\left(x^2-3x+5\right)\)
\(=-2\left(x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)
\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{11}{2}< =-\dfrac{11}{2}< 0\forall x\)
`#3107.101107`
a)
`x^2 + x + 1`
`= (x^2 + 2*x*1/2 + 1/4) + 3/4`
`= (x + 1/2)^2 + 3/4`
Vì `(x + 1/2)^2 \ge 0` `AA` `x`
`=> (x + 1/2)^2 + 3/4 \ge 3/4` `AA` `x`
Vậy, `x^2 + x + 1 > 0` `AA` `x`
b)
`4y^2 + 2y + 1`
`= [(2y)^2 + 2*2y*1/2 + 1/4] + 3/4`
`= (2y + 1/2)^2 + 3/4`
Vì `(2y + 1/2)^2 \ge 0` `AA` `y`
`=> (2y + 1/2)^2 + 3/4 \ge 3/4` `AA` `y`
Vậy, `4y^2 + 2y + 1 > 0` `AA` `y`
c)
`-2x^2 + 6x - 10`
`= -(2x^2 - 6x + 10)`
`= -2(x^2 - 3x + 5)`
`= -2[ (x^2 - 2*x*3/2 + 9/4) + 11/4]`
`= -2[ (x - 3/2)^2 + 11/4]`
`= -2(x - 3/2)^2 - 11/2`
Vì `-2(x - 3/2)^2 \le 0` `AA` `x`
`=> -2(x - 3/2)^2 - 11/2 \le 11/2` `AA` `x`
Vậy, `-2x^2 + 6x - 10 < 0` `AA `x.`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Có x2-6x+10=(x2-2.x.3+32)+1=(x-3)2+1
Vì (x-3)2 ≥0 với mọi x
nên (x-3)2+1>0 với mọi x
b) Có 4x-x2-5=-(x2-4x+4)-1=-(x2-2.x.2+22)-1=-(x-2)2-1
Vì -(x-2)2≤0 với mọi x
nên -(x-2)2-1<0 với mọi x
c)Gỉa sử (x+5)(x-3)+20>0 là đúng thì
⇔x2-3x+5x-15+20>0
⇔x2+2x+5>0 ⇔(x2+2x.1+12)+4>0 ⇔(x+1)2+4>0
Vì (x+1)2 >=0 với mọi x
Nên (x+1)2+4>0 là đúng
Vậy (x+5)(x-3)+20>0 với mọi x
a) -x2 + 6x - 10
= -(x2 - 6x + 10)
= -(x2 - 6x + 9 + 1)
= -[(x - 3)2 + 1]
Ta có: (x - 3)2 + 1 > 0 với mọi x
=> -[(x - 3)2 + 1] < 0 với mọi x
b) -2x2 - 4x - 5
= -(2x2 + 4x + 5)
= -(2x2 + 4x + 2 + 3)
= -[(\(\sqrt{2x^2}\)+\(\sqrt{2}\))2 + 3]
Ta có: (\(\sqrt{2x^2}\)+\(\sqrt{2}\))2 + 3 > 0 với mọi x
=> -[(\(\sqrt{2x^2}\)+\(\sqrt{2}\))2 + 3] < 0 với mọi x
a) \(-x^2+6x-10=-\left(x^2-6x+9\right)-1=-\left(x-3\right)^2-1< 0\forall x\)
b) \(-2x^2-4x-5=-2\left(x^2+2x+1\right)-3=-\left(x+1\right)^2-3< 0\forall x\)