Ta có: 3x = 4y => \(\frac{x}{4}=\frac{y}{3}\) => \(\frac{x}{8}=\frac{y}{6}\)

        5y = 6z => \(\frac{y}{6}=\frac{z}{5}\)

=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có: 

       \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y-z}{8+6-5}=\frac{18}{9}=2\)

=> \(\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{6}=2\\\frac{z}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.8=16\\y=2.6=12\\z=2.5=10\end{cases}}\)

Vậy ....

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+\left(z^2-6z+9\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\\z-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

\(x(3 - xy - xz) + y + 6z \le 5xz(y + z) \quad (1)\)

=>\(3x - x^2 y - x^2 z + y + 6z \le 5xyz + 5xz^2\)

=>\(3x+y+6z\le x^2y+x^2z+5xyz+5xz^2\)

=>\(3x+y+6z\le x^2(y+z)+5xz(y+z)\)

=>\(3x+y+6z\le x(y+z)(x+5z)\quad(2)\)

P=3x+y+6z

=>P<=x(y+z)(x+5z)

y+6z=P-3x

=>y+z=(P-3x)-5z

(2)=>\(P \le x(P - 3x - 5z)(x + 5z)\)

Theo AM-GM, ta có: \((P - 3x - 5z)(x + 5z) \le \left[ \frac{(P - 3x - 5z) + (x + 5z)}{2} \right]^2 = \left( \frac{P - 2x}{2} \right)^2\)

=>\(P \le x \cdot \left( \frac{P - 2x}{2} \right)^2 = \frac{x(P - 2x)^2}{4}\)

=>\(4P\le x(P^2-4Px+4x^2)\)

=>\(4P\le P^2x-4Px^2+4x^3\)

=>\(P^2x-4Px^2-4P+4x^3\ge0\)

=>\(P^2 - 4P \left(x + \frac{1}{x}\right) + 4x^2 \ge 0 \quad (3)\)

Theo Cosi, ta có: \(x + \frac{1}{x} \ge 2\sqrt{x \cdot \frac{1}{x}} = 2\)

=>\(-4P\left(x + \frac{1}{x}\right) \le -4P \cdot 2 = -8P\)

(3)=>\(P^2 - 8P + 4x^2 \ge P^2 - 4P \left(x + \frac{1}{x}\right) + 4x^2 \ge 0\)

Chọn điểm rơi x=1, ta sẽ có phương trình: \(P^2 - 8P + 4 \ge 0\)

=>\(P^2-8P+16\ge12\)

=>\(\left(P-4\right)^2\ge12\)

=>\(P-4\ge2\sqrt3\)

=>\(P\ge4+2\sqrt3\)

Dấu '=' xảy ra khi \(x=\frac{1}{x};P-3x-5z=x+5z\) và y=P-3x-6z

=>x=1; P-4x=10z và y=P-3x-6z

=>x=1; \(10z=(4+2\sqrt{3})-4=2\sqrt{3}\) và y=P-3x-6z

=>x=1; \(z=\frac{\sqrt{3}}{5}\) và y=P-3x-6z

=>\(x=1;y=1+\frac{4\sqrt{3}}{5};z=\frac{\sqrt{3}}{5}\)

1)

a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )

5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)

\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)

b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)

\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)

\(xy+yz+zx=8xyz\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=8\)

\(\Rightarrow\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}=64\)

Ta có: \(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\)

\(=\left(\dfrac{1}{x}+...+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\left(\dfrac{1}{y}+...+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}\right)+\left(\dfrac{1}{z}+...+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}\right)\)

(sau dấu chấm là bốn số tương tự).

\(\ge^{Cauchy-Schwarz}\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow64\ge\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow\dfrac{1}{6x+y+z}+\dfrac{1}{6y+z+x}+\dfrac{1}{6z+x+y}\le1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{8}\)

Vậy \(Max\) của biểu thức đã cho là 1.