Phối hợp nhiều phương pháp phân tích đa thức thành nhân tử

Ta có  ${x^2} - 7x + 10 = {x^2} - 2x - 5x + 10$$= x\left( {x - 2} \right) - 5\left( {x - 2} \right) = \left( {x - 5} \right)\left( {x - 2} \right)$

$\begin{array}{l}\,\,\,\,\,m.{n^3} - 1 + m - {n^3}\\ = \left( {m{n^3} - {n^3}} \right) + \left( {m - 1} \right)\\ = {n^3}\left( {m - 1} \right) + \left( {m - 1} \right)\\ = \left( {{n^3} + 1} \right)\left( {m - 1} \right)\\ = \left( {n + 1} \right)\left( {{n^2} - n + 1} \right)\left( {m - 1} \right).\end{array}$  

Ta có  ${x^8} + 4 = {\left( {{x^4}} \right)^2} + 4{x^4} + 4 - 4{x^4}$$= {\left( {{x^4} + 2} \right)^2} - {\left( {2{x^2}} \right)^2}$

$\begin{array}{l}\,\,\,\,\,4{x^2} + 4x - {y^2} + 1\\ = \left( {{{\left( {2x} \right)}^2} + 2.2x + 1} \right) - {y^2}\\ = {\left( {2x + 1} \right)^2} - {y^2}\\ = \left( {2x + 1 - y} \right)\left( {2x + 1 + y} \right)\\ = \left( {2x - y + 1} \right)\left( {2x + y + 1} \right).\end{array}$

Vậy đa thức trong chỗ trống là $2x - y + 1$.

Ta có:

$\begin{array}{l}{x^3} + {x^2} - 4x - 4 = \left( {{x^3} + {x^2}} \right) - \left( {4x + 4} \right)\\ = {x^2}\left( {x + 1} \right) - 4\left( {x + 1} \right) = \left( {{x^2} - 4} \right)\left( {x + 1} \right)\\ = \left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 1} \right)\end{array}$

nên A đúng.

* ${x^2} + 10x + 24$$={x^2} + 6x + 4x + 24 = x\left( {x + 6} \right) + 4\left( {x + 6} \right) = \left( {x + 4} \right)\left( {x + 6} \right)$ nên B đúng.

Vậy cả A, B đều đúng.

Ta có

+) Đáp án A đúng vì:

 $\begin{array}{l}16{x^3} - 54{y^3} = 2\left( {8{x^3} - 27{y^3}} \right)\\ = 2\left[ {{{\left( {2x} \right)}^3} - {{\left( {3y} \right)}^3}} \right]\\ = 2\left( {2x - 3y} \right)\left[ {{{\left( {2x} \right)}^2} + 2x.3y + {{\left( {3y} \right)}^2}} \right]\\ = 2\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right).\end{array}$  

+) Đáp án B đúng vì:

$\begin{array}{l}{x^2} - 9 + \left( {2x + 7} \right)\left( {3 - x} \right)\\ = \left( {{x^2} - 9} \right) + \left( {2x + 7} \right)\left( {3 - x} \right)\\ = \left( {x - 3} \right)\left( {x + 3} \right) - \left( {2x + 7} \right)\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {x + 3 - 2x - 7} \right)\\ = \left( {x - 3} \right)\left( { - x - 4} \right)\end{array}$

+) Đáp án C đúng vì:

$\begin{array}{l}\;{x^4} - 4{x^3} + 4{x^2} = {x^2}\left( {{x^2} - 4x + 4} \right)\\ = {x^2}\left( {{x^2} - 2.2.x + {2^2}} \right) = {x^2}{\left( {x - 2} \right)^2}.\end{array}$

+) Đáp án D sai vì:

$\begin{array}{l}\;4{x^3} - 4{x^2} - x + 1\\ = \left( {4{x^3} - 4{x^2}} \right) - \left( {x - 1} \right)\\ = 4{x^2}\left( {x - 1} \right) - \left( {x - 1} \right)\\ = \left( {4{x^2} - 1} \right)\left( {x - 1} \right)\\ = \left( {{{\left( {2x} \right)}^2} - 1} \right)\left( {x - 1} \right)\\ = \left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x - 1} \right).\end{array}$

Ta có  

$\begin{array}{l}\left( A \right):\;16{x^4}\left( {x - y} \right) - x + y\\ = 16{x^4}\left( {x - y} \right) - \left( {x - y} \right)\\ = \left( {16{x^4} - 1} \right)\left( {x - y} \right)\\ = \left[ {{{\left( {2x} \right)}^4} - 1} \right]\left( {x - y} \right)\\ = \left[ {{{\left( {2x} \right)}^2} - 1} \right]\left[ {\left( {2{x^2}} \right) + 1} \right]\left( {x - y} \right)\\ = \left( {2x - 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x - y} \right).\end{array}$

Nên $\left( A \right)$ sai.

Và $\left( B \right):\;2{x^3}y - 2x{y^3} - 4x{y^2} - 2xy$

 $= 2xy\left( {{x^2} - {y^2} - 2y - 1} \right) = 2xy\left[ {{x^2} - \left( {{y^2} + 2y + 1} \right)} \right]$ $= 2xy\left[ {{x^2} - {{\left( {y + 1} \right)}^2}} \right] = 2xy\left( {x - y - 1} \right)\left( {x + y + 1} \right).$

Nên $\left( B \right)$ sai.

Vậy cả (A) và (B) đều sai.

Đặt $t = {x^2} - 4x$ ta được  ${t^2} + 8t + 15 = {t^2} + 3t + 5t + 15 = t\left( {t + 3} \right) + 5\left( {t + 3} \right) = \left( {t + 5} \right)\left( {t + 3} \right)$$= \left( {{x^2} - 4x + 5} \right)\left( {{x^2} - 4x + 3} \right)$

$= \left( {{x^2} - 4x + 5} \right)\left( {{x^2} - 3x - x + 3} \right)$$= \left( {{x^2} - 4x + 5} \right)\left( {x\left( {x - 3} \right) - \left( {x - 3} \right)} \right)$$= \left( {{x^2} - 4x + 5} \right)\left( {x - 1} \right)\left( {x - 3} \right)$

Vậy số cần điền là $- 3.$

Gọi $T = (x - 1)(x - 2)(x + 4)(x + 5) - 27$$= \left[ {\left( {x - 1} \right)\left( {x + 4} \right)} \right].\left[ {\left( {x - 2} \right)\left( {x + 5} \right)} \right] - 27$$= \left( {{x^2} + 3x - 4} \right).\left( {{x^2} + 3x - 10} \right) - 27$

Đặt ${x^2} + 3x - 7 = t \Rightarrow \left\{ \begin{array}{l}{x^2} + 3x - 4 = t + 3\\{x^2} + 3x - 10 = t - 3\end{array} \right.$  từ đó ta có $T = \left( {t - 3} \right)\left( {t + 3} \right) - 27 = {t^2} - 9 - 27 = {t^2} - 36 = \left( {t - 6} \right)\left( {t + 6} \right)$

Thay $t = {x^2} + 3x - 7$ ta được $T = \left( {{x^2} + 3x - 7 - 6} \right)\left( {{x^2} + 3x - 7 + 6} \right)$$= \left( {{x^2} + 3x - 13} \right)\left( {{x^2} + 3x - 1} \right)$ suy ra $a =  - 13;b =  - 1\, \Rightarrow a + b =  - 14$

$\begin{array}{l}\,\,\,\,\,\,\,{x^3} - {x^2} - x + 1 = 0\\ \Leftrightarrow \left( {{x^3} - {x^2}} \right) - \left( {x - 1} \right) = 0\\ \Leftrightarrow {x^2}\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {{x^2} - 1} \right)\left( {x - 1} \right) = 0\end{array}$

$\begin{array}{l} \Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow {\left( {x - 1} \right)^2}\left( {x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 1 = 0\\x + 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\\x =  - 1\end{array} \right.\end{array}$

Vậy $x = 1$ hoặc $x =  - 1$.

$\begin{array}{l}\;2\left( {x + 3} \right) - {x^2} - 3x = 0\\ \Leftrightarrow 2\left( {x + 3} \right) - \left( {{x^2} + 3x} \right) = 0\\ \Leftrightarrow 2\left( {x + 3} \right) - x\left( {x + 3} \right) = 0\\ \Leftrightarrow \left( {2 - x} \right)\left( {x + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2 - x = 0\\x + 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 2\\x =  - 3\end{array} \right.\end{array}$

Vậy có hai giá trị $x$ thỏa mãn.

Ta có

$\begin{array}{l}\;{x^4} + 2{x^3} - 8x - 16 = 0\\ \Leftrightarrow \left( {{x^4} + 2{x^3}} \right) - \left( {8x + 16} \right) = 0\\ \Leftrightarrow {x^3}\left( {x + 2} \right) - 8\left( {x + 2} \right) = 0\\ \Leftrightarrow \left( {{x^3} - 8} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}{x^3} - 8 = 0\\x + 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 2\\x =  - 2\end{array} \right.\end{array}$

Mà ${x_0} < 0$ nên  ${x_0} =  - 2$ suy ra $- 3 < {x_0} <  - 1$

Ta có

$\begin{array}{l}\;{x^2} + 3x - 18 = 0\\ \Leftrightarrow {x^2} + 6x - 3x - 18 = 0\\ \Leftrightarrow \left( {{x^2} - 3x} \right) + \left( {6x - 18} \right) = 0\\ \Leftrightarrow x\left( {x - 3} \right) + 6\left( {x - 3} \right) = 0\\ \Leftrightarrow \left( {x + 6} \right)\left( {x - 3} \right) = 0\end{array}$

$\Leftrightarrow \left[ \begin{array}{l}x + 6 = 0\\x - 3 = 0\end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l}x =  - 6\\x = 3\end{array} \right.$

Suy ra ${x_1} = 3;{x_2} =  - 6\,\left( {do\,\,{x_1} > {x_2}} \right)$

$\Rightarrow \dfrac{{{x_1}}}{{{x_2}}} = \dfrac{3}{{ - 6}} =  - \dfrac{1}{2}$ .

Ta có $B = {x^3} + {x^2}y - x{y^2} - {y^3}$$= {x^2}\left( {x + y} \right) - {y^2}\left( {x + y} \right) = \left( {{x^2} - {y^2}} \right)\left( {x + y} \right)$$= \left( {x - y} \right)\left( {x + y} \right)\left( {x + y} \right) = \left( {x - y} \right){\left( {x + y} \right)^2}$

Thay $x = 3,25,\,y = 6,75$ ta được $B = \left( {3,25 - 6,75} \right){\left( {3,25 + 6,75} \right)^2} =  - 3,{5.10^2} =  - 350$ .

Ta có ${x^3} + {x^2} = 36$$\Leftrightarrow {x^3} + {x^2} - 36 = 0$

$\Leftrightarrow {x^3} - 3{x^2} + 4{x^2} - 12x + 12x - 36 = 0$ $\Leftrightarrow {x^2}\left( {x - 3} \right) + 4x\left( {x - 3} \right) + 12\left( {x - 3} \right) = 0$ $\Leftrightarrow \left( {x - 3} \right)\left( {{x^2} + 4x + 12} \right) = 0$

$\Leftrightarrow \left[ \begin{array}{l}x - 3 = 0\\{x^2} + 4x + 12 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\\{x^2} + 4x + 4 + 8 = 0\end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l}x = 3\\{\left( {x + 2} \right)^2} =  - 8\left( L \right)\end{array} \right.$

Vậy có 1 giá trị của $x$ thỏa mãn đề bài là $x = 3$ .

Ta có: $D = a\left( {{b^2} + {c^2}} \right) - b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right) - 2abc$

$= a{b^2} + a{c^2} - b{c^2} - b{a^2} + c{a^2} + c{b^2} - 2abc$ $= \left( {a{b^2} - {a^2}b} \right) + \left( {a{c^2} - b{c^2}} \right) + \left( {{a^2}c - 2abc + {b^2}c} \right)$

$= ab\left( {b - a} \right) + {c^2}\left( {a - b} \right) + c\left( {{a^2} - 2ab + {b^2}} \right)$

$=  - ab\left( {a - b} \right) + {c^2}\left( {a - b} \right) + c{\left( {a - b} \right)^2}$

$= \left( {a - b} \right)\left( { - ab + {c^2} + c\left( {a - b} \right)} \right)$

$\begin{array}{l} = \left( {a - b} \right)\left( { - ab + {c^2} + ac - bc} \right)\\ = \left( {a - b} \right)\left[ {\left( { - ab + ac} \right) + \left( {{c^2} - bc} \right)} \right]\\ = \left( {a - b} \right)\left[ {a\left( {c - b} \right) + c\left( {c - b} \right)} \right]\\ = \left( {a - b} \right)\left( {a + c} \right)\left( {c - b} \right)\end{array}$

Với $a = 99;b =  - 9;c = 1$ ,ta có:

$D = \left( {99 - \left( { - 9} \right)} \right)\left( {99 + 1} \right)\left( {1 - \left( { - 9} \right)} \right)$$= 108.100.10 = 108000$

$E = 2{x^3} - 2{y^3} - 3{x^2} - 3{y^2}$$= 2\left( {{x^3} - {y^3}} \right) - 3\left( {{x^2} + {y^2}} \right) = 2\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) - 3\left( {{x^2} + {y^2}} \right)$

Vì $x - y = 1$  nên $E = 2\left( {{x^2} + {y^2} + xy} \right) - 3{x^2} - 3{y^2}$$=  - \left( {{x^2} - 2xy + {y^2}} \right) =  - {\left( {x - y} \right)^2} =  - 1$ .

Thêm bớt $abc$ vào $M$ ta có

Ta có $M = ab\left( {a + b + c} \right) - bc\left( {b + c} \right) - abc + ca\left( {c + a} \right) + abc$ $= ab\left( {a + b + c} \right) - bc\left( {a + b + c} \right) + ac\left( {a + b + c} \right)$

$= \left( {a + b + c} \right)\left( {ab - bc + ac} \right)$

Ta có  ${x^2} - 6x + 8 = {x^2} - 4x - 2x + 8 = x\left( {x - 4} \right) - 2\left( {x - 4} \right) = \left( {x - 4} \right)\left( {x - 2} \right)$

Ta có  $25 - {a^2} + 2ab - {b^2}$$= 25 - \left( {{a^2} - 2ab + {b^2}} \right)$$= {5^2} - {\left( {a - b} \right)^2}$$= \left( {5 + a - b} \right)\left( {5 - a + b} \right)$ .