Dựa vào t/c đường trung bình bạn c/m MNPQ là hình thoi.

\(\Rightarrow NM\perp PQ\)(tính chất hình thoi.)

\(\frac{1}{x}+\frac{1}{x+6}=\frac{1}{4}\)\(\Leftrightarrow\frac{2x+6}{x\left(x+6\right)}=\frac{1}{4}\)

\(\Leftrightarrow4\left(2x+6\right)=x\left(x+6\right)\)\(\Leftrightarrow8x+24=x^2+6x\)

\(\Leftrightarrow2x+24=x^2\)\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2-2x+1-25=0\Leftrightarrow\left(x-1\right)^2=5^2\)

\(\Leftrightarrow\hept{\begin{cases}x-1=5\\x-1=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)

  Vậy \(x\in\left\{6,-4\right\}\)thì thỏa mãn đề bài

\(\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

=\(\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)

=\(\left(\frac{1}{1+\sqrt{a}}\right)^2\)

=\(\frac{1}{1+2\sqrt{a}+a}\)

\(\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)

\(=\left(\frac{1}{1+\sqrt{a}}\right)^2\)

\(=\frac{1}{1+2\sqrt{a}+a}\)

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

\(=\left(x+2\right)\left(x^2+x+1\right)\)

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

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

\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+x+1\right)\)

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

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

\(=x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(=\left(x+2\right)\left(x^2+x+1\right)\)