a) \(2x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow x^2+x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+10x+25\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x\\\left(x+5\right)^2\ge0\forall x\end{cases}}\)

\(\Rightarrow\left(x+y\right)^2+\left(x+5\right)^2\ge0\forall x\)

Vậy đẳng thức xảy ra\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=5\end{cases}}\)

b)\(x^2+3y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+y^2+2y^2+2xy-2y+\frac{1}{2}+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(2y^2-2y+\frac{1}{2}\right)+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)

Vì \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2\ge0\)

nên \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

Mà\(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)

nên pt vô nghiệm

a) 2x² + y² + 2xy + 10x + 25 = 0

=> (x² + 2xy + y²) + (x² + 10x + 25) = 0

=> (x + y)² + (x + 5)² = 0 

    <=> \(\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x\\x=-5\end{cases}}\) <=> \(\hept{\begin{cases}y=5\\x=-5\end{cases}}\)

b)c) xem lại đề

Đề sai

Mình làm câu đầu tượng trưng thui nhé, 2 câu sau tương tự vậy !!!!!!

a) pt <=> \(x^2-2xy+2y^2-2x-2y+5=0\)

<=> \(\left(x-y-1\right)^2+y^2-4y+4=0\)

<=> \(\left(x-y-1\right)^2+\left(y-2\right)^2=0\)    (1) 

TA LUÔN CÓ: \(\left(x-y-1\right)^2;\left(y-2\right)^2\ge0\forall x;y\)

=> \(\left(x-y-1\right)^2+\left(y-2\right)^2\ge0\)      (2)

TỪ (1) VÀ (2) => DẤU "=" SẼ PHẢI XẢY RA <=> \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

VẬY \(\left(x;y\right)=\left(3;2\right)\)

      2x² + 2y² -2xy+2x+2y+2=0

<=>x²-2xy+y²+x²+2x+1+y²+2y+1=0

<=>(x-y)²+(x+1)²+(y+1)²=0

<=>x=-1;y=-1

còn x²⁰¹⁶+a chia hết cho x-1 khi a =-1.đúng chuẩn

sao lại bằng -1

Bạn xem ở http://olm.vn/hoi-dap/question/106067.html

hoi-dap/question/106067.html

Vì Dũng bắt chước mình nên thấy lạ lạ chứ gì !

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Ta có: x^2+2y^2-2xy+2x+2-4y=0

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

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

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

mà (x-y+1)^2> hoặc=0 với mọi x;y

(y-1)^2> hoặc=0 với mọi x;y

nên x-y+1=0;y-1=0

=> y=1; x=0

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

\(\Leftrightarrow\left(x^2+y^2+1+2xy-2x-2y\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)^2+\left(y+1\right)^2=0\)

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

Suy ra \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=1\\y=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-1\end{cases}.}\)

\(2x^2-8x+y^2+2y+9=0\)

\(\Leftrightarrow\left(2x^2-8x+8\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+\left(y+1\right)^2=0\)

\(\Leftrightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)

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

Suy ra \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)