\(x+y+z=0\)

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

\(\Leftrightarrow\)\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow\)\(x^2+y^2+z^2=0\)   (vì  xy + yz + xz = 0)

\(\Rightarrow\)\(x=y=z=0\)

Vậy   \(Q=\left(x-1\right)^{2018}+\left(y-1\right)^{2019}+\left(z-1\right)^{2020}=1\)

Bài 1:Áp dụng C-S dạng engel

\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)

\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

Mà \(xy+yz+zx=0\)(theo đề) nên \(2\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=0\)

Vì \(\hept{\begin{cases}x^2\ge0\\y^2\ge0\\z^2\ge0\end{cases}}\) (với mọi x;y;z) nên \(x^2+y^2+z^2\ge0\) (với mọi x;y;z)

Để \(x^2+y^2+z^2=0\) \(\Leftrightarrow\) \(\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}\Leftrightarrow}x=y=z=0\)

Vậy \(A=\left(0-1\right)^{2016}+0^{2017}+\left(0+1\right)^{2018}=\left(-1\right)^{2016}+0+1^{2018}=2\)

Đk: $x\geq \frac{1}{2}$

Pt $\Leftrightarrow 4x^2+3x-7=4(\sqrt{x^3+3x^2}-2)+2(\sqrt{2x-1}-1)$

$\Leftrightarrow +4\frac{(x-1)(x+2)^2}{\sqrt{x^3+3x^2}+2}+4\frac{x-1}{\sqrt{2x-1}+1}-(x-1)(4x+7)=0$

$\Leftrightarrow (x-1)[\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-(4x+7)]=0$

$\Leftrightarrow x=1\vee \frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7=0$ $(*)$

Xét hàm số $f(x)=\frac{4(x+2)^2}{\sqrt{x^3+3x^2}+2}+\frac{4}{\sqrt{2x-1}+1}-4x-7,x\in [\frac{1}{2};+\infty )$ thì $f(x)>0,\forall x\in [\frac{1}{2};+\infty )$

$\Rightarrow $ Pt $(*)$ vô nghiệm