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

Để cho gọn, đặt \(\left(x-2y\right)^2+\left(y-3z\right)^2=a\ge0\)

\(\Rightarrow\left|x-1+a\right|=x-1-\left|\left(x-1\right)\left(2-x\right)\right|\)

- Nếu \(x-1>0\Rightarrow VT=\left|x-1+a\right|>x-1\)

Mà \(\left|\left(x-1\right)\left(2-x\right)\right|\ge0\Rightarrow VP=x-1-\left|\left(x-1\right)\left(2-x\right)\right|\le x-1\)

\(\Rightarrow VT>VP\Rightarrow\) pt vô nghiệm

- Nếu \(x-1< 0\Rightarrow VT=\left|x-1+a\right|\ge0\)

\(VP=x-1-\left|\left(x-1\right)\left(2-x\right)\right|< 0\) do \(\left\{{}\begin{matrix}x-1< 0\\\left|\left(x-1\right)\left(2-x\right)\right|\ge0\end{matrix}\right.\)

\(\Rightarrow VT>VP\Rightarrow\) pt vô nghiệm

Vậy \(x=1\), khi đó pt trở thành:

\(\left|\left(1-2y\right)^2+\left(y-3z\right)^2\right|=0\Leftrightarrow\left\{{}\begin{matrix}1-2y=0\\y-3z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\z=\dfrac{1}{6}\end{matrix}\right.\)

Vậy pt đã cho có bộ nghiệm duy nhất \(\left(x;y;z\right)=\left(1;\dfrac{1}{2};\dfrac{1}{6}\right)\)

Biện luận thiếu 1 chút rồi, ở dòng 4 có dấu "=", nên sửa từ dòng 4 đến dòng 6 bằng đoạn này:

\(x-1>0\Rightarrow VT=\left|x-1+a\right|\ge x-1\)

\(VP=x-1-\left|\left(x-1\right)\left(2-x\right)\right|\le x-1\)

\(\Rightarrow VT\ge VP\), dấu "=" xảy ra khi và chỉ khi:

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

Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

\(-A=2x^2+y^2+2xy-3x-2y-2\)

\(-A=\left(x^2+2xy+y^2\right)+x^2-4x-2y-2\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)-4\)

\(-A=\left(x+y-1\right)^2+\left(x-1\right)^2-4\)

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

\(B=\left(x-2y+5\right)^2+\left(y-1\right)^2+1\)

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

       \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

Khó quá do thi phuong nhung

\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(M=2x^2+9y^2-6xy-6x-12y+2028=\left(x+2\right)^2-6y\left(x+2\right)+9y^2+\left(x-5\right)^2+1999=\left(x+2-3y\right)^2+\left(x-5\right)^2+2019\ge1999\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(N=x^2-4xy+5y^2+10x-22y+28=\left(x+5\right)^2-4y\left(x+5\right)+4y^2+\left(y-1\right)^2+2=\left(x+5-2y\right)^2+\left(y-1\right)^2+2\ge2\)

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

Mình đang bận nên chỉ nói hướng làm thôi nhá. GTNN thì bạn cộng trừ 1, còn GTLN thì bạn cộng trừ 6. Sau đó bạn sẽ tách ra được thành a+(2x^2+y^2)/x^2+y^2 

hjvbm 

\(C=x^2-4xy+5y^2+10x-22y+28\)

\(=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\)

\(=\left(x-2y-5\right)^2+\left(y-1\right)^2+2\ge2\)

Đẳng thức khó tìm quá huhu