\[\boxed{\text{711.}\text{\ }еуроки - ответы\ на\ пятёрку}\]
Пояснение.
Решение.
\[\textbf{а)}\ n(n - 1) - (n + 3)(n + 2) =\]
\[= n^{2} - n - \left( n^{2} + 2n + 3n + 6 \right) =\]
\[= n^{2} - n - n^{2} - 5n - 6 =\]
\[= - 6n - 6 =\]
\[= - 6 \cdot (n + 1) \Longrightarrow делится\ на\ 6\]
\[при\ любом\ значении\ n.\]
\[\textbf{б)}\ n(n + 2) - (n - 7)(n - 5) =\]
\[= n^{2} + 2n - \left( n^{2} - 5n - 7n + 35 \right) =\]
\[= 14n - 35 = 7 \cdot (2n - 5) \Longrightarrow\]
\[делится\ на\ 7\ при\ любом\ \]
\[значении\ n.\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\text{711\ (711).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{3} + x^{2} + x + 1 =\]
\[= \left( x^{3} + x^{2} \right) + 1 \cdot (x + 1) =\]
\[= x^{2}(x + 1) + x + 1 =\]
\[= (x^{2} + 1)(x + 1)\]
\[\textbf{б)}\ y^{5} - y^{3} - y^{2} + 1 =\]
\[= \left( y^{5} - y^{3} \right) - \left( y^{2} - 1 \right) =\]
\[= y^{3}\left( y^{2} - 1 \right) - {(y}^{2} - 1) =\]
\[= (y^{2} - 1)(y^{3} - 1)\]
\[\textbf{в)}\ a^{4} + 2a^{3} - a - 2 =\]
\[= \left( a^{4} - a \right) + \left( 2a^{3} - 2 \right) =\]
\[= a\left( a^{3} - 1 \right) + 2 \cdot \left( a^{3} - 1 \right) =\]
\[= (a^{3} - 1)(a + 2)\]
\[\textbf{г)}\ b^{6} - 3b^{4} - 2b^{2} + 6 =\]
\[= \left( b^{6} - 2b^{4} \right) - \left( 3b^{4} - 6 \right) =\]
\[= b^{2}\left( b^{4} - 2 \right) - 3 \cdot \left( b^{4} - 2 \right) =\]
\[= (b^{4} - 2)(b^{2} - 3)\]
\[\textbf{д)}\ a^{2} - ab - 8a + 8b =\]
\[= \left( a^{2} - 8a \right) - (ab - 8b) =\]
\[= a(a - 8) - b(a - 8) =\]
\[= (a - 8)(a - b)\]
\[\textbf{е)}\ ab - 3b + b^{2} - 3a =\]
\[= (ab - 3a) + \left( b^{2} - 3b \right) =\]
\[= a(b - 3) + b(b - 3) =\]
\[= (b - 3)(a + b)\]
\[\textbf{ж)}\ 11x - xy + 11y - x^{2} =\]
\[= (11x + 11y) - \left( x^{2} + xy \right) =\]
\[= 11 \cdot (x + y) - x(x + y) =\]
\[= (x + y)(11 - x)\]
\[\textbf{з)}\ kn - mn - n^{2} + mk =\]
\[= \left( kn - n^{2} \right) + (mk - mn) =\]
\[= n(k - n) + m(k - n) =\]
\[= (k - n)(n + m)\ \]