C = -3x² - 6x - 12

    = -3( x² + 2x + 1 ) - 9

    = -3( x + 1 )² - 9 ≤ -9 < 0 ∀ x ( đpcm )

D = -4x² - 12x - 15

     = -4( x² + 3x + 9/4 ) - 6

     = -4( x + 3/2 )² - 6 ≤ -6 < 0 ∀ x ( đpcm )

E = -30 - 5x² + 10x

    = -5( x² - 2x + 1 ) - 25

    = -5( x - 1 )² - 25 ≤ -25 < 0 ∀ x ( đpcm )

\(C=-3x^2-6x-12\)

\(\Rightarrow C=-\left(3x^2+6x+12\right)\)

\(\Rightarrow C=-\left(3x^2+6x+3+9\right)\)

\(\Rightarrow C=-\left[3\left(x+1\right)^2+9\right]\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2+9\ge9\)

\(\Rightarrow C=-\left[3\left(x+1\right)^2+9\right]\le-9\)

=> Đpcm

\(D=-4x^2-12x-15\)

\(\Rightarrow D=-\left(4x^2+12x+15\right)\)

\(\Rightarrow D=-\left[4\left(x+\frac{3}{2}\right)^2+6\right]\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow4\left(x+\frac{3}{2}\right)^2+6\ge6\)

\(\Rightarrow D=-\left[4\left(x+\frac{3}{2}\right)^2+6\right]\le-6\)

=> Đpcm

\(E=-30-5x^2+10x\)

\(\Rightarrow E=-\left(5x^2-10x+30\right)\)

\(\Rightarrow E=-\left[5\left(x-1\right)^2+25\right]\)

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow5\left(x-1\right)^2+25\ge25\)

\(\Rightarrow E=-\left[5\left(x-1\right)^2+25\right]\le-25\)

=> Đpcm

Ta có BĐT sau:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

CM:    \(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)

<=>   \(a^2+b^2+c^2-ab-bc-ca\ge0\)

<=>   \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)     (*)

=> BĐT (*) LUÔN ĐÚNG !!!!

=>   \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)

=>   \(3\left(ab+bc+ca\right)\le0\)

=>   \(ab+bc+ca\le0\)

VẬY TA CÓ ĐPCM.

\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+ac+ca\right)=0\)

Vì  \(a^2+b^2+c^2\ge0\forall a;b;c\)

\(\Rightarrow2\left(ab+bc+ca\right)\le0\)

\(\Rightarrow ab+bc+ca\le0\left(đpcm\right)\)

\(B=-10-x^2-6x\)

\(\Rightarrow B=-\left(x^2+6x+10\right)\)

\(\Rightarrow B=-\left(x^2+6x+9+1\right)\)

\(\Rightarrow B=-\left[\left(x+3\right)^2+1\right]\)

Vì \(\left(x+3\right)^2\ge0\forall x\)\(\Rightarrow\left(x+3\right)^2+1\ge1\)

\(\Rightarrow-\left[\left(x+3\right)^2+1\right]\le-1\)

=> Đpcm

B=\(-10-x^2-6x\)  

B=\(-x^2-6x-9-1\) 

B=\(-\left(x^2+6x+9\right)-1\)    

=\(-\left(x+3\right)^2-1\)   

Ta có : \(\left(x+3\right)^2\ge0\forall x\) 

\(-\left(x+3\right)^2\le0\) 

\(-\left(x+3\right)^2-1\le-1\)      

Vậy B luôn âm với mọi x 

(x+5)³-x³-125

=x³+5³-x³-5³

=0

1. \(\left(x+5\right)^3-x^3-125\)

\(=x^3+15x^2+75x+125-x^3-125\)

\(=15x^2+75x\)

2. \(x^3+6x^2+12x+8=0\)

\(\Leftrightarrow x^3+2x^2+4x^2+8x+4x+8=0\)

\(\Leftrightarrow x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)^2\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)^3=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

5x² - 5xy + 4y - 4x

= 5x ( x - y ) - 4 ( x - y )

= ( 5x - 4 ) ( x - y )

( x + y )³ + ( x - y )³

= 2x³ + 6xy²

= 2x ( x² + 3y² )

5 x^2 - 5xy + 4y - 4x 

= 5x ( x - y ) - 4 ( x - y ) 

= ( x - y ) ( 5x - 4 ) 

( x + y )^3 + ( x - y )^3 

= \(x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\) 

= \(2x^3+6xy^2\) 

=\(2x\left(x^2+3y^2\right)\) 

a) \(5x\left(x-1\right)-3x\left(x-1\right)\)

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

b) \(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

c) \(x\left(x-y\right)+y\left(y-x\right)\)

\(=\left(x-y\right)\left(x-y\right)\)

\(=\left(x-y\right)^2\)

d) \(x^2+xy+x=x\left(x+y+1\right)\)

a. 5x ( x - 1 ) - 3x ( x - 1 )

= ( 5x - 3x ) ( x - 1 )

b. x ( x + y ) - 5x - 5y

= x ( x + y ) - 5 ( x + y )

= ( x - 5 ) ( x + y )

c. x ( x - y ) + y ( y - x )

= x ( x - y ) - y ( x - y )

= ( x - y )²

 d. x² + xy + x 

= x ( x + y + 1 )