\(\frac{x}{2}-\frac{1}{y}=\frac{2}{3}\)

<=>\(\frac{x}{2}-\frac{2}{3}=\frac{1}{y}\)

<=>\(\frac{3x-4}{6}=\frac{1}{y}\)

<=>\(y\left(3x-4\right)=6\)

Ta có bảng sau:

Vì x;y là số nguyên nên có 2 cặp số nguyên thỏa mãn là ..........

2.

a.

\(x^2+3x=k^2\)

\(\Leftrightarrow4x^2+12x=4k^2\)

\(\Leftrightarrow4x^2+12x+9=4k^2+9\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(2k\right)^2+9\)

\(\Leftrightarrow\left(2x+3\right)^2-\left(2k\right)^2=9\)

\(\Leftrightarrow\left(2x+3-2k\right)\left(2x+3+2k\right)=9\)

Vậy \(x=\left\{-4;-3;0;1\right\}\)

b. Tương tự

\(x^2+x+6=k^2\)

\(\Leftrightarrow4x^2+4x+24=4k^2\)

\(\Leftrightarrow\left(2k\right)^2-\left(2x+1\right)^2=23\)

\(\Leftrightarrow\left(2k-2x-1\right)\left(2k+2x+1\right)=23\)

Em tự lập bảng tương tự câu trên

1.

\(\Leftrightarrow x^2-2xy+y^2=-4y^2+y+1\)

\(\Leftrightarrow-4y^2+y+1=\left(x-y\right)^2\ge0\)

\(\Leftrightarrow-64y^2+16y+16\ge0\)

\(\Leftrightarrow\left(8y-1\right)^2\le17\)

\(\Rightarrow\left(8y-1\right)^2\le16\)

\(\Rightarrow-4\le8y-1\le4\)

\(\Rightarrow-\dfrac{3}{8}\le y\le\dfrac{5}{8}\)

\(\Rightarrow y=0\)

Thế vào pt ban đầu:

\(\Rightarrow x^2=1\Rightarrow x=\pm1\)

Vậy \(\left(x;y\right)=\left(-1;0\right);\left(1;0\right)\)

Từ giả thiết ta có :

\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

ta có : \(Q=\frac{y+2}{x^2}+\frac{z+2}{y^2}+\frac{x+2}{z^2}\)

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

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

\(\ge\frac{2\left(x+1\right)}{zx}+\frac{2\left(y+1\right)}{xy}+\frac{2\left(z+1\right)}{yz}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2\)

Áp dụng bđt \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Dấu " = " xảy ra khi và chỉ khi a = b = c

Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\ge3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=3\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\sqrt{3}\)

Do đó : \(Q\ge\sqrt{3}+2\). Dấu " = " xảy ra 

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\\z+y+z=xyz\end{cases}\Leftrightarrow x=y=z=\sqrt{3}}\)

Vậy Min \(Q=\sqrt{3}+2\)khi \(x=y=z=\sqrt{3}\)