Ta có:

\(x^2+4y^2+z^2-4x+4y-8z+24=0\)

\(\Leftrightarrow x^2-4x+4+4y^2+4y+1+z^2-8z+16+3=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(4y^2+4y+1\right)+\left(z^2-8z+16\right)+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+3=0\)

Mà: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2y+1\right)^2\ge0\\\left(z-4\right)^2\ge0\end{matrix}\right.\)

 \(\Rightarrow\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+3\ge3\ne0\)

Vậy không có số thực x, y, z nào thỏa mãn đẳng thức.

x² - 3y² + 2xy + 2x - 4y - 7 = 0

<=> 4.(x² - 3y² + 2xy + 2x - 4y - 7) = 0

<=> 4x² - 12y² + 8xy + 8x - 16y - 28 = 0

<=> (4x² + 8xy + 4y²) + (8x + 8y) + 4 - 16y² - 24y - 32 = 0

<=> (2x + 2y)² + 4(2x + 2y) + 4 - (16y² + 24y + 9) = 23

<=> (2x + 2y + 2)² - (4y + 3)² = 23

<=> (2x + 6y + 5)(2x - 2y - 1) = 23

Vì \(x;y\inℤ\Rightarrow2x+6y+5;2x-2y-1\inℤ\) 

Lập bảng : 

Vậy (x;y) = (3;2) ; (-9;2) 

\(a,\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\right]\\ b,\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\\ \Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\\ \Leftrightarrow x,y,z\in\varnothing\left[\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\right]\)

\(c,\Leftrightarrow-\left(x^2-10xy+25y^2\right)-\left(y^2-20y+100\right)-50=0\\ \Leftrightarrow-\left(x-5y\right)^2-\left(y-10\right)^2-50=0\\ \Leftrightarrow x,y\in\varnothing\left[-\left(x-5y\right)^2-\left(y-10\right)^2-50\le-50< 0\right]\)

a/

\(\Leftrightarrow x^2-2xy+y^2+2x^2+10x+26=0\)

\(\Leftrightarrow\left(x-y\right)^2+2\left(x-\frac{5}{2}\right)^2+\frac{27}{2}=0\)

\(VT>0\Rightarrow\) ko tồn tại x; y thỏa mãn

b/

\(\Leftrightarrow4x^2-4x+1+3\left(y^2+10y+25\right)+2=0\)

\(\Leftrightarrow\left(2x-1\right)^2+3\left(y+5\right)^2+2=0\)

\(\Rightarrow\) Không tồn tại x; y thỏa mãn

c/

\(3\left(x^2-4x+4\right)+6\left(y^2-\frac{10}{3}y+\frac{25}{9}\right)+\frac{34}{3}=0\)

\(\Leftrightarrow3\left(x-2\right)+6\left(y-\frac{5}{3}\right)^2+\frac{34}{3}=0\)

Không tồn tại x; y thỏa mãn

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

a: \(x^2+3y^2-4x+6y+7=0\)

\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)

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

\(4x^2+8y^2-8xy+4x-8y+4=0\)

\(4x^2-4x\left(2y-1\right)+\left(2y-1\right)^2+8y^2-8y+4-\left(2y-1\right)^2=0\)

\(\left(2x-2y+1\right)^2+\left(4y^2-4y+1\right)+3=0\)

\(\left(2x-2y+1\right)^2+\left(2y-1\right)^2+3=0\) ( vô lí)

=> KL...........

vô lí

Lần này là lần thứ 3 tớ gửi câu này