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\(P=\sqrt{\dfrac{x^3}{y+3}}+\sqrt{\dfrac{y^3}{z+3}}+\sqrt{\dfrac{z^3}{x+3}}\)
\(=\dfrac{x^2}{\sqrt{x\left(y+3\right)}}+\dfrac{y^2}{\sqrt{y\left(z+3\right)}}+\dfrac{z^2}{\sqrt{z\left(x+3\right)}}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}}\)
Xét:
\(\left(\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\right)^2\le\left(1^2+1^2+1^2\right)\left(xy+3x+yz+3y+xz+3z\right)\)
\(=3\left(9+xy+yz+xz\right)\)
\(=27+3\left(xy+yz+xz\right)\le27+\left(x+y+z\right)^2=36\)
\(\Rightarrow\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\le6\)
\(P\ge\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=1\)

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{2\cdot2+3\cdot3-4}=\dfrac{45}{9}=5\)
Do đó: x-1=10; y-2=15; z-3=20
=>x=11; y=17; z=23
c: Ta có: 10x=6y
nên x/3=y/5
Đặt x/3=y/5=k
=>x=3k; y=5k
Ta có: \(2x^2-y^2=-28\)
\(\Leftrightarrow2\cdot9k^2-25k^2=-28\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
=>x=6; y=10
TRường hợp 2: k=-2
=>x=-6; y=-10

\(A=\dfrac{x+y}{z}+1+\dfrac{x+z}{y}+1+\dfrac{y+z}{x}+1-3\)
\(A=\left(x+y+z\right)\left(\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)-3\)
\(A=0-3=-3\)

\(P=\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{z^2}{2}+\dfrac{xy+xz+yz}{xyz}\)
\(\Rightarrow P\ge\dfrac{x^2}{2}+\dfrac{1}{x}+\dfrac{y^2}{2}+\dfrac{1}{y}+\dfrac{z^2}{2}+\dfrac{1}{z}\)
\(\Rightarrow P\ge\left(\dfrac{x^2}{2}+\dfrac{1}{2x}+\dfrac{1}{2x}\right)+\left(\dfrac{y^2}{2}+\dfrac{1}{2y}+\dfrac{1}{2y}\right)+\left(\dfrac{z^2}{2}+\dfrac{1}{2z}+\dfrac{1}{2z}\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{x^2}{2}.\dfrac{1}{2x}.\dfrac{1}{2x}}+3\sqrt[3]{\dfrac{y^2}{2}.\dfrac{1}{2y}.\dfrac{1}{2y}}+3\sqrt[3]{\dfrac{z^2}{2}.\dfrac{1}{2z}.\dfrac{1}{2z}}=\dfrac{9}{2}\)
\(\Rightarrow P_{min}=\dfrac{9}{2}\) khi \(x=y=z=1\)