CMR: \(x^3+4x+1>3x^2\) với x ≥ 0
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DL
2
HJ
15 tháng 5 2016
\(Q=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(Q=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)+\left(x-1\right)\)
\(Q=\left(x-1\right)\left(x^3-2x^2+2x+1\right)_{\ge}0\)
15 tháng 5 2016
\(Q=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(Q=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)+\left(x-1\right)\)
\(Q=\left(x-1\right)\left(x^3-2x^2+2x+1\right)\ge0\)
D
4
10 tháng 6 2017
a , Ta có \(x^2+x+1=x^2+2x\frac{1}{2}+\left(\frac{1}{2}\right)^2+\)\(\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\) \(\ge\frac{3}{4}>0\left(đpcm\right)\)
b , Ta có : \(4x^2-2x+3\)= \(\left(2x\right)^2-2.2x.1+1^2+2\) = \(\left(2x-1\right)^2+2\ge2>0\left(đpcm\right)\)
c , Ta có \(3x^2+2x+1=x^2-\frac{2x}{3}+\frac{1}{9}+2x^2+\frac{8x}{3}+\frac{8}{9}\)
= \(\left(x-\frac{1}{3}\right)^2+2\left(x^2+\frac{4x}{3}+\frac{4}{9}\right)=\left(x-\frac{1}{3}\right)^2+2\left(x+\frac{2}{3}\right)^2\ge0\)
Vì Dấu "=" không thể xảy ra , do đó \(3x^2+2x+1>0\left(đpcm\right)\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 10 2023
a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))
\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)
\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)
b) Thay x = 2003 ta có:
\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)
c) \(C>0\) khi:
\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\) (đpcm)
A
30 tháng 6 2021
Bài 1
\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)
\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)
Bài 2
\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)
1 tháng 3 2018
có : \(x\ge0\)
\(\Rightarrow x^3+4x\ge0\)
\(\Rightarrow\)\(x^3+4x+1\ge1\)
có \(3x^2\ge0\) ( vì x >=0)
suy ra
\(x^3+4x+1\ge3x^2\)
ND
2
13 tháng 7 2017
a ) \(4x^2+2x+1=\left(2x\right)^2+2\cdot2x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
b ) \(x^2+3x+4=\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)
c ) \(9x^2+3x+5=\left(3x\right)^2+2\cdot3x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}>0\forall x\)
13 tháng 7 2017
Ta có : 4x2 + 2x + 1
= (2x)2 + 2.2x.\(\frac{1}{2}\) + \(\frac{1}{2}+\frac{3}{4}\)
= (2x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\)
Mà : (2x + \(\frac{1}{2}\))2 \(\ge0\forall x\)
=> (2x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\)
Hay : (2x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) \(>0\forall x\)
Vậy 4x2 + 2x + 1 \(>0\forall x\)
TN
0
TN
1
DN
21 tháng 8 2018
Ta có:\(-x^2+4x-7\)
\(=-\left(x^2-4x+7\right)\)
\(=-\left(x^2-2.x.2+2^2-4+7\right)\)
\(=-\left[\left(x-2\right)^2+3\right]\)
\(=-\left(x-2\right)^2-3\)
Do \(-\left(x-2\right)^2\le0\) với \(\forall x\)
\(\Rightarrow-\left(x-2\right)^2-3\le-3< 0\)
\(\Rightarrow-x^2+4x-7< 0\) (đpcm)
câu b,c đề sai bạn nhé!
TL
Cho đa thức: \(f\left(x\right)=4x^3+4x^4-x^2+3x^2-3x^4-3x^3\). CMR f(x) chỉ có 1 nghiệm x=0
Giúp hộ!
1
29 tháng 3 2018
\(f\left(x\right)=4x^3+4x^4-x^2+3x^2-3x^4-3x^3\)
\(\Leftrightarrow f\left(x\right)=\left(4x^3-3x^3\right)+\left(4x^4-3x^4\right)+\left(-x^2+3x^2\right)\)
\(\Leftrightarrow f\left(x\right)=x^3+x^4+2x^2\)
\(f\left(x\right)=0\)
\(\Leftrightarrow x^3+x^4+2x^2=0\)
\(\Leftrightarrow x^2\left(x+x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{3}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\forall x\end{matrix}\right.\)
Vậy f(x) chỉ có 1 nghiệm