\(P=\frac{x^4}{x^2y^2+x^2yz+z^2x^2}+\frac{y^4}{y^2z^2+xzy^2+x^2y^2}+\frac{z^4}{z^2x^2+xyz^2+y^2z^2}\)

ÁP DỤNG BĐT CAUCHY -  SCHWARZ TA ĐƯỢC:

=>   \(P\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2y^2+y^2z^2+z^2x^2\right)+xyz\left(x+y+z\right)}\)           (1)

TA SẼ CHỨNG MINH:    \(\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2y^2+y^2z^2+z^2x^2\right)+xyz\left(x+y+z\right)}\ge1\)         (2)

<=>   \(x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge2\left(x^2y^2+y^2z^2+z^2x^2\right)+xyz\left(x+y+z\right)\)

<=>   \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)        (*)

TA ÁP DỤNG LIÊN TỤC 2 LẦN DẠNG BĐT SAU:     \(\alpha^2+\beta^2+\gamma^2\ge\alpha\beta+\beta\gamma+\alpha\gamma\)

KHI ĐÓ TA SẼ ĐƯỢC:    \(\Rightarrow x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)\)

VẬY BĐT (*) LÀ LUÔN ĐÚNG.

=> TỪ (1) VÀ (2)    =>    \(P\ge1\)

DẤU "=" XẢY RA <=>    \(x=y=z\)

VẬY P MIN = 1 <=>    x = y = z .

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

=>    \(\hept{\begin{cases}\frac{x}{4}=\frac{y}{6}\\\frac{y}{6}=\frac{z}{9}\end{cases}}\)

=>    \(\frac{x}{4}=\frac{y}{6}=\frac{z}{9}\)

=>    \(\frac{x}{4}=\frac{y}{6}=\frac{z}{9}=\frac{x+y+z}{4+6+9}=\frac{38}{19}=2\)

=>   \(\frac{x}{4}=2;\frac{y}{6}=2;\frac{z}{9}=2\)

=>    \(x=8;y=12;z=18.\)

Ta có \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{2}=\frac{z}{3}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{4}=\frac{y}{6}\\\frac{y}{6}=\frac{z}{9}\end{cases}}\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{9}\)

Lại có x + y + z = 38

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{9}=\frac{x+y+z}{4+6+9}=\frac{38}{19}=2\)

=> x = 8 ; y = 12 ; z = 18

BÀI 1:

a) 

PT <=>    \(3x-2=7-4\sqrt{3}\)

<=>    \(3x=9-4\sqrt{3}\)

<=>    \(x=3-\frac{4}{\sqrt{3}}\)

b)

pt =>   \(x+1=14-6\sqrt{5}\)

<=>   \(x=13-6\sqrt{5}\)

BÀI 2: 

a)

pt <=>   \(\sqrt{x^2-9}=3\sqrt{x-3}\)

<=>   \(x^2-9=9\left(x-3\right)\)

<=>   \(x^2-9=9x-27\)

<=>   \(x^2-9x+18=0\)

<=>   \(\orbr{\begin{cases}x=6\\x=3\end{cases}}\)

BÀI 2: 

b)

pt <=>   \(\sqrt{x^2-4}=2\sqrt{x+2}\)

<=>   \(x^2-4=4\left(x+2\right)\)

<=>   \(x^2-4=4x+8\)

<=>   \(x^2-4x-12=0\)

<=>   \(\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

BÀI 3:

pt <=>   \(x^2=5\)

<=>   \(\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=2\)

\(\Rightarrow\frac{43}{8}+x-\frac{173}{24}=\frac{100}{3}\)

\(\Rightarrow x-\frac{11}{6}=\frac{100}{3}\)

\(\Rightarrow x=\frac{211}{6}\)

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