Cmr \(3x^2-x+1\)>0 vs mọi x
cmr 3x^2+4x-2 < 0 vs mọi x
Cmr
a)-x+4x-7<0 với mọi x
b)-x^2-10x-3<0, với mọi x
c)-x^2+3x-4<0, vs mọi x
Cmr
a)-x+4x-7<0 với mọi x
b)-x^2-10x-3<0, với mọi x
c)-x^2+3x-4<0, vs mọi x
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!
CMR
a) x^2-7x+16>0 vs moi x thuoc R
b) 3x^2-3x+1>0 vs moi x thuoc R
c) -x^2+3x-5<0 vs moi x thuoc R
d) 3x-2x^2<0 vs moi x thuoc R
CMR:
a) x^2-7x+16>0 vs moi x thuoc R
b) 3x^2-3x+1>0 vs moi x thuoc R
c) -x^2+3x-5<0 vs moi x thuoc R
d) 3x-2x^2<0 vs moi x thuoc R
a) x2 - 7x + 16
= (x2 - 2x\(\frac{7}{2}\)+ \(\frac{49}{4}\)) + \(\frac{15}{4}\)
= (x - \(\frac{7}{2}\))2 + \(\frac{15}{4}\)> 0
b) 3x2 - 3x + 1
= [\(\left(\sqrt{3x^2}\right)^2\)- 2.\(\sqrt{3x^2}\).\(\frac{\sqrt{3}}{2}\)+ \(\frac{3}{4}\)] + \(\frac{1}{4}\)
= (\(\sqrt{3x^2}\)- \(\frac{\sqrt{3}}{2}\))2 + \(\frac{1}{4}\)> 0
c) -x2 + 3x - 5
= -(x2 - 3x + 5)
= -(x2 - 2x\(\frac{3}{2}\)+ \(\frac{9}{4}\)+\(\frac{11}{4}\))
= -[(x - \(\frac{3}{2}\))2 + \(\frac{11}{4}\)] < 0
d) Câu này sai đề rồi bạn ơi
1. CMR
a) x^2-3x+5>0 với mọi x
b) x^2-1/3x+5/4>0 với mọi x
c)x-x^2-3<0 với mọi x
d) x-2x^2-5/2<0 với mọi x
hơi ngán dạng này :((((
a, \(x^2-3x+5=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)
b,
\(x^2-\frac{1}{3}x+\frac{5}{4}=x^2-2.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{5}{4}=\left(x-\frac{1}{6}\right)^2+\frac{11}{9}>0\forall x\)
c,
\(x-x^2-3=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}-3=-\left(x-\frac{1}{2}\right)^2-\frac{11}{4}< 0\forall x\)d,
\(x-2x^2-\frac{5}{2}=-2\left(x^2-\frac{1}{2}x+\frac{5}{4}\right)=-2\left(x^2-2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{4}\right)=-2\left[\left(x-\frac{1}{4}\right)^2+\frac{19}{16}\right]=-2\left(x-\frac{1}{4}\right)^2-\frac{19}{8}< 0\forall x\)P/s : ko chắc lém :)))
CMR
a) 4x2+2x+1>0 với mọi x
b)x2+3x+4>0 với mọi x
c)9x2+3x+5>0 với mọi x
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 ) \(9x^2+3x+5=\left(3x\right)^2+2\cdot3x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}>0\forall x\)
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 \(>0\forall x\)
CMR: Q = x^4 - 3x^3 + 4x^2 - 3x +1 >= 0 với mọi x
\(Q=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(Q=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)+\left(x-1\right)\)
\(Q=\left(x-1\right)\left(x^3-2x^2+2x+1\right)_{\ge}0\)
\(Q=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(Q=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)+\left(x-1\right)\)
\(Q=\left(x-1\right)\left(x^3-2x^2+2x+1\right)\ge0\)
Bài 1:CMR:
a, x2+x+1>0 với mọi x
b, 4x2-2x+3>0 với mọi x
c, 3x2+2x+1 >0 với mọi x
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 \(3x^2+2x+1>0\left(đpcm\right)\)