X^2 - 4x + 1 < 0
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YF
0
NC
23 tháng 8 2020
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) \(E=-4x^2+4x-2=-\left(2x-1\right)^2-1\le-1< 0\left(\forall x\right)\)
KN
23 tháng 8 2020
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
16 tháng 7 2021
| 2-4x | = 4x-2
<=> \(\orbr{\begin{cases}\left|2-4x\right|=-2+4x=4x-2\\\left|2-4x\right|=2-4x=4x-2\end{cases}}\)
<=>\(\orbr{\begin{cases}-2+4x=4x-2\\2-4x=4x-2\end{cases}}\)
<=>\(\orbr{\begin{cases}-2+4x-4x+2=0\\2-4x-4x+2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}0=0\\-8x+4=0\end{cases}}\)
<=> x=\(\frac{-4}{-8}=\frac{1}{2}\)
=> \(S=\left\{\frac{1}{2};\infty\right\}\)
2x-7> 3(x-1)
<=>2x-7>3x-3
<=>2x-3x>-3+7
<=>-x>4
<=>x<4
=>S={x/x<4}
1-2x<4(3x-2)
<=>1-2x<12x-8
<=>-2x-12x<-8-1
<=>-14x<-9
<=>x>\(\frac{9}{14}\)
=>S={\(\frac{9}{14}\)}
-3x+2|-4 -x|> 0
<=>\(\orbr{\begin{cases}-3x+2+4+x>0\\-3x+2-4x-x>0\end{cases}}\)
<=>\(\orbr{\begin{cases}-2x+6>0\\-8x+2>0\end{cases}}\)
<=>\(\orbr{\begin{cases}-2x>-6\\-8x>-2\end{cases}}\)
<=>\(\orbr{\begin{cases}x< 3\\x< \frac{1}{4}\end{cases}}\)
=>S={x/x<3;x/x<\(\frac{1}{4}\)}
4x-1|x-2|< 0
<=>\(\orbr{\begin{cases}4x-1-x+2< 0\\4x-1+x-2< 0\end{cases}}\)
<=>\(\orbr{\begin{cases}3x+1< 0\\3x-3< 0\end{cases}}\)
<=>\(\orbr{\begin{cases}3x< -1\\3x< 3\end{cases}}\)
<=>\(\orbr{\begin{cases}x< \frac{-1}{3}\\x< 1\end{cases}}\)
=>S={x/x<\(\frac{-1}{3}\);x/x<1}
DK
1
9 tháng 8 2017
a, \(A=-5x^2+10x-7=-5\left(x^2-2x+1\right)^2-2=-5\left(x-1\right)^2-2< 0\)
\(\Rightarrowđpcm\)
b, \(B=-x^2+x-\dfrac{1}{4}\)
\(=-\left(x^2-\dfrac{1}{2}.x.2+\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2\le0\)
c, \(C=-4x^2+4x-3=-\left(4x^2-4x+1+2\right)\)
\(=-\left(2x-1\right)^2-2< 0\)
\(\Rightarrowđpcm\)
TT
0
AT
13 tháng 6 2018
1. 4x2 + 4x + 2 = (4x2 + 4x + 1) + 1 = (2x + 1)2 + 1
Có: (2x+1)2 ≥ 0 ∀x => (2x+1)2 + 1 ≥ 1 > 0 (đpcm)
3. -x2 + 4x - 5 = -(x2 - 4x + 4) - 1 = -(x - 2)^2 - 1
Có: -(x-2)^2 ≤ 0 => -(x-2)^2 -1 ≤ - 1 < 0 (đpcm)
7. (x+2)(x-5) + 15 = x2 - 3x + 5 = (x2 - 2.x.\(\dfrac{3}{2}\)+ \(\dfrac{9}{4}\)) + \(\dfrac{11}{4}\)
= ( x - \(\dfrac{3}{2}\))^2 + \(\dfrac{11}{4}\) \(\ge\dfrac{11}{4}>0\left(đpcm\right)\)
KS
0
AT
14 tháng 6 2018
2. \(-x^2+2x-2=-\left(x^2+2x+1\right)-1=-\left(x+1\right)^2-1\)
vì: \(-\left(x+1\right)^2\forall x\le0\Rightarrow-\left(x+1\right)^2-1\le-1< 0\left(đpcm\right)\)
6.
\(\left(x-2\right)\left(x-4\right)+3=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\)
vì: \(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+2\ge2>0\left(đpcm\right)\)
Giải BPT?
Ta có: \(x^2-4x+1< 0\)
\(\Leftrightarrow\left(x^2-4x+4\right)-3< 0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{3}\right)^2< 0\)
\(\Leftrightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)< 0\)
Nhận thấy \(x-2-\sqrt{3}< x-2+\sqrt{3}\)
=> \(\hept{\begin{cases}x-2-\sqrt{3}< 0\\x-2+\sqrt{3}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2+\sqrt{3}\\x>2-\sqrt{3}\end{cases}}\)
Vậy \(2-\sqrt{3}< x< 2+\sqrt{3}\)