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a,2x2+8x+20=2(x2+4x)+20
=2(x2+4x+4)+20-4.2
=2(x+2)2+12
Ta có : 2(x+2)2 \(\ge0với\forall x\)
12 > 0
\(\Rightarrow\)2(x+2)2+12>0 với \(\forall x\)
\(\Rightarrow\)2x2+8x+20>0 với \(\forall\)x
b,x4-3x2+5
=(x4-3x2)+5
=(x4-2.\(\frac{3}{2}\)x2+\(\frac{9}{4}\))+5-\(\frac{9}{4}\)
=(x2-\(\frac{3}{2}\))2+\(\frac{11}{4}\)
Có : (x2-3/2)2\(\ge0với\forall x\)
\(\frac{11}{4}\)>0
\(\Rightarrow\)(x2-\(\frac{3}{2}\))2+\(\frac{11}{4}>0với\forall x\)
a. \(x^2+3x+5\)
\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
=> đpcm
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 \(>0\forall x\)
Bài 1.
a) ( 7x - 3 )2 - 5x( 9x + 2 ) - 4x2 = 18
<=> 49x2 - 42x + 9 - 45x2 - 10x - 4x2 = 18
<=> -52x + 9 = 18
<=> -52x = 9
<=> x = -9/52
b) ( x - 7 )2 - 9( x + 4 )2 = 0
<=> x2 - 14x + 49 - 9( x2 + 8x + 16 ) = 0
<=> x2 - 14x + 49 - 9x2 - 72x - 144 = 0
<=> -8x2 - 86x - 95 = 0
<=> -8x2 - 10x - 76x - 95 = 0
<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0
<=> ( x + 5/4 )( -8x - 76 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)
c) ( 2x + 1 )2 + ( 4x - 1 )( x + 5 ) = 36
<=> 4x2 + 4x + 1 + 4x2 + 19x - 5 = 36
<=> 8x2 + 23x - 4 - 36 = 0
<=> 8x2 + 23x - 40 = 0
=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))
Bài 2.
a) x2 - 12x + 39 = ( x2 - 12x + 36 ) + 3 = ( x - 6 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
b) 17 - 8x + x2 = ( x2 - 8x + 16 ) + 1 = ( x - 4 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
c) -x2 + 6x - 11 = -( x2 - 6x + 9 ) - 2 = -( x - 3 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
d) -x2 + 18x - 83 = -( x2 - 18x + 81 ) - 2 = -( x - 9 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
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
a)Ta có: x2+x+1
=x2+2.x.1/2+1/4+3/4
=(x+1/2)2+3/4
Vì (x+1/2)2>=0 với mọi x
=>(x+1/2)2+3/4>0 với mọi x
Vậy x2+x+1>0 với mọi x.
b)Ta có: -5-x2+2x
=-(x2-2x+5)
=-(x2-2x+1+4)
=-(x-1)2-4
Ta có:(x-1)2>=0 với mọi x
=>-(x-1)2<=0 với mọi x
=>-(x-1)2-4<0 với mọi x
Vậy -5-x2+2x<0 với mọi x
a) x2+x+1 = \(x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)
= \(x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{3}{4}\)
=\(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Do \(\left(x+\frac{1}{2}\right)^2\le0\)vs mọi x => \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)vs mọi x
=> x^2 + x + 1 > 0 vs mọi x
b) -5-x^2 + 2x = -(x^2 - 2x + 5) = \(-\left(x^2-2x+1+4\right)=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\)
Do \(-\left(x-1\right)^2\le0\)vs mọi x=> \(-\left(x-1\right)^2-4< 0\)vs mọi x
=> -5-x^2+2x<0 vs 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 :)))
cảm ơn bạn nhìuuu 💞