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

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

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

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

=>    \(x=8;y=12;z=18.\)

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 

GTLN chứ ?

\(P\le\frac{1}{9}\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}+\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}+\frac{1}{az}+\frac{1}{bx}+\frac{1}{cy}\right)\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

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tìm giá trị nhỏ nhất cơ mà bạn PHÙNG MINH QUÂN ???