ĐKXĐ: x;y khác 0

Áp dụng bđt AM-GM cho 2 số dương ta có:

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}\)\(\ge2\sqrt{x^2.\frac{1}{x^2}}+2\sqrt{y^2.\frac{1}{y^2}}=2+2=4\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix}x^2=\frac{1}{x^2}\\y^2=\frac{1}{y^2}\end{matrix}\right.\)<=>\(\left\{\begin{matrix}x^4=1\\y^4=1\end{matrix}\right.\)<=>\(\left\{\begin{matrix}\left[\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy (x;y)=(1;1) ; (x;y)=(1;-1) ; (x;y)=(-1;1) ; x;y = (-1;-1)

x^2 + y^2 + 1/x^2 + 1/y^2 -4 =0

<=> x^2 + 1/x^2 -2 + y^2 +1/y^2 -2 = 0

<=>x^2 -2 + 1/x^2 + y^2 -2 +1/y^2= 0

<=> (x-1/x)^2 +(y-1/y)^2 =0

=> phương trình vô nghiệm do bình phương luôn luôn dương nên hai bình phương cộng lại không thể bằng 0.

<=>x^2-2+1/x^2+y^2-2+1/y^2=0

<=>(x-1/x)^2+(y-1/y)^2=0

<=>x=1/x va y=1/y

=>(x;y) thuoc cong tru 1

vay......

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

\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y-4=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=6\)

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

\(\Rightarrow\left(x-1\right)^2\le6\forall x\)

\(\Rightarrow-\sqrt{6}\le x-1\le\sqrt{6}\)

\(\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)

Từ đó thay vào tìm các giá trị tương ứng của y.

b) (x+1)(x+7)(x+3)(x+5)+15=0

=> (x^2+7x+x+7)(x^2+5x+3x+15)+15=0

=> (x^2+8x+7)(x^2+8x+15)+15=0

bạn chơi roblox à

\(x^4+x^2-y^2-y+20=0\)

<=> x²(x²+1)-y(y+1)=-20

\(\left\{{}\begin{matrix}3x+y=3\\3x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=0\\3x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}6x=0\\3x-y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3.0-y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

\(2x^2+3xy+y^2=0\)

\(\Rightarrow2x^2+2xy+xy+y^2=0\)

\(\Rightarrow2x\left(x+y\right)+y\left(x+y\right)=0\)

\(\Rightarrow\left(x+y\right)\left(2x+y\right)=0\)

     \(2x^2+3xy+y^2=0\)

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

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

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

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

Hoặc \(x+y=0\Leftrightarrow x=-y\left(1\right)\)

Hoặc \(2x+y=0\left(2\right)\)

Thế (1) vào (2) ta có: 

\(-2y+y=0\)

\(\Leftrightarrow-y=0\Leftrightarrow y=0\)

\(\Leftrightarrow x=0\left(\text{vì x = -y}\right)\)

Vậy \(x=y=0\)

Ta có : \(2x^2+3xy+y^2=2x^2+2xy+xy+y^2=2x\left(x+y\right)+y\left(x+y\right)=\left(2x+y\right)\left(x+y\right)=0\)

\(=>\orbr{\begin{cases}2x+y=0\\x+y=0\end{cases}=>\orbr{\begin{cases}x=-\frac{y}{2}\\x=-y\end{cases}}}\)

Vậy x=-y hoặc x=-y/2 với mọi x thì 2x^2+3xy+y^2