Ta có:

\(x^2+x^2y^2-2y=0\)

\(\Leftrightarrow x^2=\frac{2y}{y^2+1}\le1\)(cái này chứng minh đơn giản b tự làm lấy nhé)

\(\Leftrightarrow-1\le x\le1\left(1\right)\)

Ta lại có:

\(x^3+2y^2-4y+3=0\)

\(\Leftrightarrow x^3=-1-2\left(y-1\right)^2\le-1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x=-1\)

\(\Rightarrow y=1\)

\(\Rightarrow x^2+y^2=1+1=2\)

kdfjeuy;r;

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3x^2+3y^2+4xy-2x+2y+2=0

=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0

=>x=1 và y=-1

M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\2x^2+2x^2y^2-4y=0\left(1\right)\end{matrix}\right.\Rightarrow}x^3+2y^2-4y-2x^2-2x^2y^2+4y=0\Rightarrow x^3+1-2x^2y^2+2y^2-2x^2+2=0\Rightarrow\left(x+1\right)\left(x^2-x+1\right)-2y^2\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(x^2-x+1-2xy^2+2y^2-2x+2\right)=0\Rightarrow x=-1\)Thay x=-1 vào (1) ta được y²-2y+1=0⇒ (y-1)²=0⇒y-1=0⇒y=1

Do đó Q=x²+y²=(-1)²+1²=2

Đặt \(x=a;2y=b;3z=c\Rightarrow a+b+c=3\) 

\(T=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)

Áp dụng Bđt Cô si ngược dấu ta có:

\(T=\text{∑}a-\frac{a^2b}{1+b^2}\ge\text{∑}a-\frac{a^2b}{2b}=\text{∑}a-\frac{ab}{2}\)

\(=a+b+c-\frac{ab+bc+ca}{2}\ge a+b+c-\frac{\left(ab+bc+ca\right)^2}{6}\)\(=3-\frac{3^2}{6}=\frac{3}{2}\)

Dấu = khi \(a=b=c=1\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=\frac{1}{3}\end{cases}}\)