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a) \(x^2-x+1\)
\(=\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
b) \(x^2+2x+2\)
\(=\left(x^2+2x+1\right)+1\)
\(=\left(x+1\right)^2+1>0\forall x\)
c) \(-x^2+4x-5\)
\(=-x^2+4x-4-1\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1< 0\forall x\)
1)
a) \(3x^3y^2-6x^2y^3+9x^2y^2\)
\(=3x^2y^2\left(x-2y+3\right)\)
b) \(5x^2y^3-25x^3y^4+10x^3y^3\)
\(=5x^2y^3\left(1-5xy+2x\right)\)
Bài 1:
a)-x^2+4x-5
=-(x2-4x+5)<0 với mọi x
=>-x^2+4x-5<0 với mọi x
b)x^4+3x^2+3
\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x
=>x^4+3x^2+3>0 với mọi x
c) bn xét từng th ra
Bài 2:
a)9x^2-6x-3=0
=>3(3x2-2x-1)=0
=>3x2-2x-1=0
=>3x2+x-3x-1=0
=>x(3x+1)-(3x+1)=0
=>(x-1)(3x+1)=0
b)x^3+9x^2+27x+19=0
=>(x+1)(x2+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)
- Với x+1=0 =>x=-1
- Với x2+8x+19 =>vô nghiệm
c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
=>x3-25x-x3-8=3
=>-25x-8=3
=>-25x=1
=>x=-11/25
= (x2-x+1)(x2+3x+10)+10 = P
x2-x+1=(x-\(\frac{1}{2}\))2+\(\frac{3}{4}\)>0
x2+3x+10=(x+\(\frac{3}{2}\))2+\(\frac{31}{4}\)>0
vây P>0
Ta có :
2x4 + 1 - 2x3 - x2
= 2x3 ( x - 1 ) - ( x - 1 ) ( x + 1 )
= ( x - 1 ) ( 2x3 - x - 1 )
= ( x - 1 ) [ ( x3 - x ) + ( x3 - 1 ) ]
= ( x - 1 ) [ x ( x - 1 ) ( x + 1 ) + ( x - 1 ) ( x2 + x + 1 ) ]
= ( x - 1 )2 ( x2 + x + x2 + x + 1 )
= ( x - 1 )2 ( 2x2 + 2x + 1 )
= ( x - 1 )2 ( x2 + ( x + 1 )2 ) \(\ge\)0
Suy ra đpcm
Bài làm:
a) Ta có: \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\left(\forall x\right)\)
=> đpcm
b) \(x^4+3x^2+3=\left(x^4+3x^2+\frac{9}{4}\right)+\frac{3}{4}=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)
=> đpcm
a) -x2 + 4x - 5 = -x2 + 4x - 4 - 1
= -( x2 - 4x + 4 ) - 1
= -( x - 2 )2 - 1 ≤ -1 < 0 ∀ x ( đpcm )
b) x4 + 3x2 + 3 ( * )
Đặt t = x2
(*) <=> t2 + 3t + 3
<=> ( t2 + 3t + 9/4 ) + 3/4
<=> ( t + 3/2 )2 + 3/4
<=> ( x2 + 3/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )
a) A=x4 +3x2+3
A=(x2)2+2.\(\dfrac{3}{2}\) x2+\(\left(\dfrac{3}{2}\right)^2\) +\(\dfrac{3}{4}\)
A=(x4+3x2+\(\dfrac{9}{4}\) )+\(\dfrac{3}{4}\)
A=\(\left(x^2+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\)
do \(\left(x^2+\dfrac{3}{2}\right)^2\ge0\forall x\)
=>\(\left(x^2+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
=>A≥\(\dfrac{3}{4}\)
vậy A >1(đpcm)