\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

Bạn xem lại phương trình ban đầu có đúng không vậy?

2x²+y²+9=6x+2xy

=>2x²+y²+9-6x-2xy=0

=>(x²-2xy+y²)+(x²-6x+9)=0

=>(x-y)²+(x-3)²=0

do (x-y)² ≥ 0 ∀ x,y

(x-3)² ≥ 0 ∀x

=>(x-y)²+(x-3)² =0 khi

=>\(\left[{}\begin{matrix}x-y=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=x=3\\x=3\end{matrix}\right.\)

thay x=3 và y=3

Q=3²⁰¹⁷.3²⁰¹⁸-3²⁰¹⁸. 3²⁰¹⁷+\(\dfrac{1}{9}.3.3\)

Q=1

Lời giải:
Ta có:

\(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)

\(\Leftrightarrow (4x^2-4xy+y^2)+2z^2+y^2-2z(2x-y)-6y-10z+34=0\)

\(\Leftrightarrow (2x-y)^2-2z(2x-y)+z^2+(y^2-6y+9)+(z^2-10z+25)=0\)

\(\Leftrightarrow (2x-y-z)^2+(y-3)^2+(z-5)^2=0\)

Vì \((2x-y-z)^2; (y-3)^2; (z-5)^2\geq 0, \forall x,y,z\). Do đó để \((2x-y-z)^2+(y-3)^2+(z-5)^2=0\) thì:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\)

\(\Rightarrow \left\{\begin{matrix} x=4\\ y=3\\ z=5\end{matrix}\right.\)

Khi đó:

\(S=(4-4)^{2018}+(3-4)^{2019}+(5-4)^{2020}=0+(-1)+1=0\)

Đặt biểu thức trên là A, thay xyz = 2018, ta dược :

\(A=\dfrac{x^2yz}{xy+xyz+x^2yz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+x+1}\)

\(=\dfrac{xy\left(xz\right)}{xy\left(1+z+xz\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{z+zx+1}\)

\(=\dfrac{xz}{1+z+xz}+\dfrac{1}{z+1+xz}+\dfrac{z}{z+zx+1}=\dfrac{xz+1+z}{1+z+xz}=1\)

⇒ĐPCM

Please help me!!!!!!!!!!!

I feel this exercise is difficult!!!!!!