\(x+y+z=0\)

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

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

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

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

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

Ta có: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\)

=>\(\frac{yz + zx + xy}{xyz}=0\)

=>xy+yz+xz=0

=>yz=-xy-xz; xy=-xz-yz; xz=-xy-yz

\(x^2 + 2yz = x^2 + yz + yz = x^2 - x(y + z) + yz\)

\(=x^2-xy-xz+yz\)

\(=x(x-y)-z(x-y)=(x-y)(x-z)\)

Chứng minh tương tự, ta sẽ có:

\(y^2 + 2zx = (y - z)(y - x)\)

\(z^2 + 2xy = (z - x)(z - y)\)

Đặt \(A = \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy}\)

\(=\frac{1}{(x - y)(x - z)}+\frac{1}{(y - z)(y - x)}+\frac{1}{(z - x)(z - y)}\)

\(=\frac{-1}{(x - y)(z - x)}+\frac{-1}{(y - z)(x - y)}+\frac{-1}{(z - x)(y - z)}\)

\(=\frac{-(y - z) - (z - x) - (x - y)}{(x - y)(y - z)(z - x)}\)

=0

=>\(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}=0\)

\(\left(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)\)

\(=0\left(x^{2016}+y^{2017}+z^{2018}\right)\)

=0

=xy+yz+xz

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

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

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