CMR: \(x-x^2-1< 0\forall x\)
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4 tháng 9 2018
-x^ - x - 1 = - (x^2+x+1) = - (x^2+x+1/4+3/4) = - [(x+1/2)^2 +3/4) ]
Ta có [(X+1/2)^2+3/4 lớn hơn hoặc bằng 3/4 => - [(x+1/2)^2+3/4] nhỏ hơn hoặc bằng -3/4 <0
4 tháng 9 2018
\(-\left(x^2+x+1\right)\Rightarrow-\left[x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2\right]\)
\(\Rightarrow-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\Rightarrow-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\Rightarrow\le0\)
NT
1
14 tháng 8 2022
\(-x^2+x-\dfrac{1}{2}\)
\(=-\left(x^2-x+\dfrac{1}{2}\right)\)
\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{1}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}< 0\)
NT
3
16 tháng 8 2018
a ) \(x^2+4x+5=x^2+2.x.2+2^2+1=\left(x+2\right)^2+1\)
\(Do\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1>0\forall x\left(đpcm\right)\)
b) \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
\(Do\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\left(đpcm\right)\)
c)\(-\left(4x^2-12x+9\right)-1=-\left(2x-3\right)^2-1\)
\(Do-\left(2x-3\right)\le0\Rightarrow-\left(2x-3\right)-1\le-1\forall x\)
16 tháng 8 2018
\(x^2+2.x.2+2^2+5-4\) \(\Rightarrow\left(x+2\right)^2+5-4\) \(\Rightarrow\left(x+2\right)^2+1\)
vì \(\left(x+2\right)^2\ge0\) \(\Rightarrow\left(x+2\right)^2+1\ge1\) \(\ge0\) \(\Rightarrow dpcm\)
b) \(x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2\) \(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{5}{4}\)
vì \(\left(x+\frac{1}{2}\right)^2\ge0\) \(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\ge0\) \(\Rightarrow dpcm\)
c) \(12x-4x^2-10=-\left(4x^2-12x+10\right)\) = \(\left[\left(2x\right)^2-2.2x.3+3^2\right]+10-3^2\)
\(\Rightarrow\left(2x-3\right)^2+10-9\) \(\Rightarrow\left(2x-3\right)^2+1\) vì \(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+1\ge1hay\ge0\left(1>0\right)\Rightarrow dpcm\)
.
, ta có: 
.![(3-x)(2+x) \leq \dfrac{25}{4}, \, \forall x\in [-2; 3]](https://s0.wp.com/latex.php?latex=%283-x%29%282%2Bx%29+%5Cleq+%5Cdfrac%7B25%7D%7B4%7D%2C+%5C%2C+%5Cforall+x%5Cin+%5B-2%3B+3%5D&bg=ffffff&fg=4e4e4e&s=0 "(3-x)(2+x) \leq \dfrac{25}{4}, \, \forall x\in [-2; 3]")
.![(2-x).(1+2x) \leq \dfrac{25}{8}, \, \forall x\in [-\dfrac{1}{2}; 2]](https://s0.wp.com/latex.php?latex=%282-x%29.%281%2B2x%29+%5Cleq+%5Cdfrac%7B25%7D%7B8%7D%2C+%5C%2C+%5Cforall+x%5Cin+%5B-%5Cdfrac%7B1%7D%7B2%7D%3B+2%5D&bg=ffffff&fg=4e4e4e&s=0 "(2-x).(1+2x) \leq \dfrac{25}{8}, \, \forall x\in [-\dfrac{1}{2}; 2]")
TM
0
\(x-x^2-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}-1\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)
Ta có :
\(-\left(x-\dfrac{1}{2}\right)^2\le0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}< 0\forall x\)
hay \(x-x^2-1< 0\forall x\)