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

\(1.A=x^2+2x+2=\left(x+1\right)^2+1\)

Vì\(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1\forall x\)

hay\(\left(x+1\right)^2+1>0\forall x\)

\(2.B=x^2+2x.3+9+2=\left(x+3\right)^2+2\)

CM tương tự A

\(3.C=4x^2+4x-2=\left(2x+1\right)^2-2\)

Vì\(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+1\right)^2-2\ge-2\forall x\)(có thể >0)

4,5 Cm tương tự

Ta có : x² + 2x + 2

= x² + 2x + 1 + 1

= (x + 1)² + 1 \(\ge1\forall x\)

Vậy  x² + 2x + 2 \(>0\forall x\)

Ta có : x² + 2x + 2

=> x² + 2x + 1 + 1

=> ( x + 1)² + 1  >  1\(\forall x\)

Vậy x² + 2x + 2   > \(0\forall x\)

a) Xét \(x^2-4x+4=\left(x-2\right)^2\ge0\)

<=> \(x^2-4x\ge-4>-5\)

b) \(2x^2+4y^2-4x-4xy+5\)

= \(\left(x^2-4x+4\right)+\left(x^2-4xy+4y^2\right)+1\)

= \(\left(x-2\right)^2+\left(x-2y\right)^2+1\ge1>0\)

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 )

Bài làm:

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

\(=-\left(2x+1\right)^2-1\le-1< 0\left(\forall x\right)\)

=> đpcm

b) \(x^2+4y^2+z^2-2x-6z+8y+15\)

\(=\left(x^2-2x+1\right)+\left(4y^2-8y+4\right)+\left(z^2-6z+9\right)+1\)

\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

                                           \(=-\left(2x+1\right)^2-1\)

    Vì \(-\left(2x+1\right)^2\le0\forall x\)\(\Rightarrow\)\(-\left(2x+1\right)^2-1\le-1\forall x\)

              \(\Rightarrow\)\(-\left(2x+1\right)^2-1< 0\forall x\)

              \(\Rightarrow\)\(-4x^2-4x-2< 0\forall x\)( ĐPCM )

b) Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)

        \(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)

        \(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)

    Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\)\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\forall x,y,z\)( ĐPCM )

a) Ta có : -4x² - 4x - 2 = -(4x² + 4x + 1) - 1 = -(2x + 1)² - 1 < 0 (đpcm)

b) x² + 4y² + z² - 2x - 6z + 8y + 15

= (x² - 2x + 1) + (z² - 6z + 9) + (4y² + 8y + 4) + 1

= (x - 1)² + (z - 3)² + 4(y + 1)² + 1 > 0 (đpcm)

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

b. \(4x^2+5x+7\)

\(=\left(2x\right)^2-2.2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{87}{16}\)

= \(\left(2x+\dfrac{5}{4}\right)^2\) + \(\dfrac{87}{16}\) \(\ge\dfrac{87}{16}\)

=> đpcm

c. \(6x-9x^2-4\)

=> \(-9x^2+6x-4\)

\(=-\left(9x^2-6x+4\right)\)

\(=-[\left(3x\right)^2-2.3x.1+1-5]\)

\(=\left(3x-1\right)^2\le5\forall x\)

=> đpcm

Bài 1:

\(a,A=2x^2+2x+1=\left(x^2+2x+1\right)+x^2=\left(x+1\right)^2+x^2\\ Mà:\left(x+1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x+1\right)^2+x^2>0\forall x\in R\\ Vậy:A>0\forall x\in R\)

2:

a: =-(x^2-3x+1)

=-(x^2-3x+9/4-5/4)

=-(x-3/2)^2+5/4 chưa chắc <0 đâu bạn

b: =-2(x^2+3/2x+3/2)

=-2(x^2+2*x*3/4+9/16+15/16)

=-2(x+3/4)^2-15/8<0 với mọi x

Bài 1:

\(B=4+x^2+x=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\\ Vậy:B>0\forall x\in R\)

A) x²+4y²2+z²2-4x-6z+15>0 <=> (x²-2×2×x+2²)+4y²+(z²-2×3×z+3²) +(15 -2²-3²) >0

<=>(x-2)²+4y²2+(z-3)²

B) giải

(2X)²+ 2×2X×1 +1 >=0 với mọi X (   (2x+1)²  )

=> (2x+1)²+2 >0