Trắc nghiệm Bài 4: Vận dụng hằng đẳng thức vào phân tích đa thức thành nhân tử Toán 8 Cánh diều

Ta có:

\(\begin{array}{*{20}{l}}{2{x^2}\;-4x + 2 = 0}\\{\;2\left( {{x^2}\;-2x + 1} \right) = 0}\\{\;2{{\left( {x-1} \right)}^2}\; = 0}\\{\;x-1 = 0}\\{\;x = 1}\end{array}\)

Vậy x = 1

\(\begin{array}{*{20}{l}}{4{b^2}{c^2}\;-{{\left( {{c^2}\; + {b^2}\;-{a^2}} \right)}^2}}\\{\;\;\;\;\;\;\;\;\; = {{\left( {2bc} \right)}^2}\;-{{\left( {{c^2}\; + {b^2}\;-{a^2}} \right)}^2}}\\{\;\;\;\;\;\;\;\;\; = \left( {2bc + {c^2}\; + {b^2}\;-{a^2}} \right)\left( {2bc-{c^2}\;-{b^2}\; + {a^2}} \right)}\\{\;\;\;\;\;\;\;\;\; = \left[ {{{\left( {b + c} \right)}^2}\;-{a^2}} \right]\left[ {{a^2}\;-\left( {{b^2}\;-2bc + {c^2}} \right)} \right]}\\{\;\;\;\;\;\;\;\;\; = \left[ {{{\left( {b + c} \right)}^2}\;-{a^2}} \right]\left[ {{a^2}\;-{{\left( {b-c} \right)}^2}} \right]}\\{\;\;\;\;\;\;\;\;\; = \left( {b + c + a} \right)\left( {b + c-a} \right)\left( {a + b-c} \right)\left( {a-b + c} \right)}\end{array}\)

\({x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\)

Ta có

\(P = {x^3}\;-3{x^2}\; + 3x-1 + 1 = {\left( {x-1} \right)^3}\; + 1\)

Thay x = 1001 vào P ta được

\(P = {\left( {1001-1} \right)^3}\; + 1 = {1000^3}\; + 1\)

\(\begin{array}{*{20}{l}}{2 - 25{x^2} = 0\;}\\{ \Leftrightarrow (\sqrt 2 - 5x)(\sqrt 2 + 5x) = 0}\\\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\sqrt 2 - 5x = 0\\\sqrt 2 + 5x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{\sqrt 2 }}{5}\\x = \frac{{ - \sqrt 2 }}{5}\end{array} \right.\end{array}\end{array}\)

Ta có:

\(\begin{array}{*{20}{l}}{{x^6}\;-{y^6}\; = {{\left( {{x^3}} \right)}^2}\;-{{\left( {{y^3}} \right)}^2}\; = \left( {{x^3}\; + {y^3}} \right)\left( {{x^3}\;-{y^3}} \right)}\\{ = \left( {x + y} \right)\left( {{x^2}\;-xy + {y^2}} \right)\left( {x-y} \right)\left( {{x^2}\; + xy + {y^2}} \right)}\\{}\end{array}\)

\(\begin{array}{l}{37^2} - {13^2}\\ = \left( {37 - 13} \right)\left( {37 + 13} \right)\\ = 24.50\\ = 1200\end{array}\)

\({x^2} - 2xy + {y^2}{\rm{ - }}81\; = \;\left( {{x^2} - 2xy + {y^2}} \right) - 81\) (nhóm 3 hạng tử đầu để xuất hiện bình phương một hiệu)

\( = {\left( {x - y} \right)^2} - {9^2}\) (áp dụng hằng đẳng thức \({A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)\))

\( = \left( {x - y - 9} \right)\left( {x - y + 9} \right)\).

\({x^2} + 2x + 1 - {y^2} = \left( {{x^2} + 2x + 1} \right) - {y^2}\;\) (nhóm hạng tử)

\( = {\left( {x + 1} \right)^2} - {y^2}\) (áp dụng hằng đẳng thức)

\( = \left( {x + 1 - y} \right)\left( {x + 1 + y} \right)\)

Thay x = 94,5 và y = 4,5 vào biểu thức, ta được:

\(\begin{array}{*{20}{l}}{\left( {{\rm{94}},{\rm{5}} + 1 - 4,5} \right)\left( {{\rm{94}},{\rm{5}} + 1 + {\rm{4}},{\rm{5}}} \right)}\\{ = 91.100}\\{ = 9100}\end{array}\)

+) \({x^2} - 6x + 9 = {x^2} - 2.3x + {3^2} = {(x - 3)^2}\) nên A đúng.

+) \(\frac{{{x^2}}}{4} + 2xy + 4{y^2} = {\left( {\frac{x}{2}} \right)^2}.2.\frac{x}{2}.2y + {\left( {2y} \right)^2} = {\left( {\frac{x}{2} + 2y} \right)^2}\) nên B sai, C đúng.

+) \(4{x^2} - 4xy + {y^2} = {\left( {2x} \right)^2} - 2.2x.y + {y^2} = {(2x - y)^2}\) nên D đúng.

\(\begin{array}{l}{\left( {3{x^2} + 3x - 5} \right)^2} - {\left( {3{x^2} + 3x + 5} \right)^2}\\ = (3{x^2} + 3x - 5 - 3{x^2} - 3x - 5)(3{x^2} + 3x - 5 + 3{x^2} + 3x + 5\\ = - 10(6{x^2} + 6x)\\ = - 10.6x(x + 1)\\ = - 60x(x + 1)\\ = mx(x + 1)\\ \Rightarrow m = - 60 < 0\end{array}\)

\(\begin{array}{l}A = {x^4} + 3{x^3} - 27x - 81\\ = ({x^4} - 81) + (3{x^3} - 27x)\\ = ({x^2} - 9)({x^2} + 9) + 3x({x^2} - 9)\\ = ({x^2} - 9)({x^2} + 3x + 9)\end{array}\)

Ta có: \({x^2} + 3x + 9 = {x^2} + 2.\frac{3}{2}x + \frac{9}{4} + \frac{{27}}{4} \ge \frac{{27}}{4} > 0,\forall x\)

Mà \(\left| x \right| < 3 \Leftrightarrow {x^2} < 9 \Leftrightarrow {x^2} - 9 < 0\)

\( \Rightarrow A = ({x^2} - 9)({x^2} + 3x + 9) < 0\) khi \(\left| x \right| < 3\).

\(\begin{array}{l}{(3{x^2} + 6x - 18)^2} - {(3{x^2} + 6x)^2}\\ = (3{x^2} + 6x - 18 - 3{x^2} - 6x)(3{x^2} + 6x - 18 + 3{x^2} + 6x)\\ = - 18(6{x^2} + 12x - 18)\\ = - 18.6({x^2} + 2x - 3)\\ = - 108({x^2} + 2x - 3)\\ = - 108({x^2} - x + 3x - 3)\\ = - 108\left[ {x(x - 1) + 3(x - 1)} \right]\\ = - 108(x + 3)(x - 1)\end{array}\)

Khi đó, m = -108; n = 3 \( \Rightarrow \frac{m}{n} = \frac{{ - 108}}{3} = - 36\)

\(\begin{array}{*{20}{l}}{B = {x^3}\; + 3{x^2}y + 3x{y^2}\; + {y^3}\; + {x^2}\; + 2xy + {y^2}}\\{ = \left( {{x^3}\; + 3{x^2}y + 3x{y^2}\; + {y^3}} \right) + \left( {{x^2}\; + 2xy + {y^2}} \right)}\\{ = {{\left( {x + y} \right)}^3}\; + {{\left( {x + y} \right)}^2}\; = {{\left( {x + y} \right)}^2}\left( {x + y + 1} \right)}\end{array}\)

Vì \(x = 20-y\) nên \(x + y = 20\). Thay \(x + y = 20\) vào \(B = {\left( {x + y} \right)^2}\left( {x + y + 1} \right)\) ta được:

\(B = {\left( {20} \right)^2}\left( {{\rm{20 }} + 1} \right) = 400.21 = 8400\).

Vậy \(B > 8300\) khi \(x = 20-y\).

Gọi hai số lẻ liên tiếp là \(2k-1;2k + 1(k \in N*)\)

Theo bài ra ta có:

\({\left( {2k + 1} \right)^{2}}-{\left( {2k-1} \right)^{2}} = 4{k^2} + 4k + 1-4{k^2} + 4k-1 = 8k \vdots 8,\forall k \in \mathbb{N}*\)

\(\begin{array}{l}5{x^2} - 10x + 5 = 0\\ \Leftrightarrow 5({x^2} - 2x + 1) = 0\\ \Leftrightarrow {(x - 1)^2} = 0\\ \Leftrightarrow x - 1 = 0\\ \Leftrightarrow x = 1\end{array}\)

\(\begin{array}{*{20}{l}}{{{\left( {2x-5} \right)}^2}\;-4{{\left( {x-2} \right)}^2}\; = 0}\\{ \Leftrightarrow {{\left( {2x-5} \right)}^2}\;-{{\left[ {2\left( {x-2} \right)} \right]}^2}\; = 0}\\{ \Leftrightarrow {{\left( {2x-5} \right)}^2}\;-{{\left( {2x-4} \right)}^2}\; = 0}\\{ \Leftrightarrow \left( {2x-5 + 2x-4} \right)\left( {2x-5-2x + 4} \right) = 0}\\{ \Leftrightarrow \left( {4x-9} \right).\left( { - 1} \right) = 0}\\{ \Leftrightarrow - 4x + 9 = 0}\\{ \Leftrightarrow 4x = 9}\\{ \Leftrightarrow x = \;\frac{9}{4}}\end{array}\)

Ta có:

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}{x^3}\; + {x^2}\;-4x-4\\ = \left( {{x^3}\; + {x^2}} \right)-\left( {4x + 4} \right)\end{array}\\\begin{array}{l} = {x^2}\left( {x + 1} \right)-4\left( {x + 1} \right)\\ = \left( {{x^2}\;-4} \right)\left( {x + 1} \right)\end{array}\end{array}\\ = \left( {x-2} \right)\left( {x + 2} \right)\left( {x + 1} \right)\end{array}\)

nên A đúng.

\(\begin{array}{l}{x^2}\; + 10x + 24\\ = {x^2}\; + 6x + 4x + 24\\ = x\left( {x + 6} \right) + 4\left( {x + 6} \right)\\ = \left( {x + 4} \right)\left( {x + 6} \right)\end{array}\)

nên B đúng.

Vậy cả A, B đều đúng

Ta thấy a + b + c = 0 nên \({a^3} + {b^3} + {c^3} = 3abc\).

\(\begin{array}{l}4{\left( {2x-5} \right)^2}\;-9{(4{x^2}\;-25)^2}\; = 0\\\begin{array}{*{20}{l}}{ \Leftrightarrow 4{{\left( {2x-5} \right)}^2}\;-9{{[{{\left( {2x} \right)}^2}\;-{5^2}]}^2}\; = 0}\\{ \Leftrightarrow 4{{\left( {2x-5} \right)}^2}\;-9{{\left[ {\left( {2x-5} \right)\left( {2x + 5} \right)} \right]}^2}\; = 0}\\{ \Leftrightarrow 4{{\left( {2x-5} \right)}^2}\;-9{{\left( {{\rm{2x }}-5} \right)}^2}{{\left( {2x + 5} \right)}^2}\; = 0}\\{ \Leftrightarrow {{\left( {2x-5} \right)}^2}[4-9{{\left( {2x + 5} \right)}^2}] = 0}\\{ \Leftrightarrow {{\left( {2x-5} \right)}^2}[4-{{\left( {3\left( {2x + 5} \right)} \right)}^2}] = 0}\\{ \Leftrightarrow {{\left( {2x-5} \right)}^2}({2^2}\;-{{\left( {6x + 15} \right)}^2}) = 0}\\{ \Leftrightarrow {{\left( {3x-5} \right)}^2}\left( {2 + {\rm{ 6}}x + 15} \right)\left( {2-{\rm{ 6}}x-15} \right) = 0}\\\begin{array}{l} \Leftrightarrow {\left( {3x-5} \right)^2}\left( {6x + 17} \right)\left( { - 6x-13} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{5}{3}\\x = \frac{{ - 17}}{6}\\x = \frac{{13}}{6}\end{array} \right.\end{array}\end{array}\end{array}\)

Suy ra \({x_1} + {x_2} + {x_3} = \frac{5}{3} - \frac{{17}}{6} + \frac{{13}}{6} = \frac{{10 - 17 + 13}}{6} = 1\)

Từ đẳng thức đã cho suy ra \({a^3}\; + {b^3}\; + {c^3}\;-3abc = 0\)

\(\begin{array}{*{20}{l}}{{b^3}\; + {c^3}\; = \left( {b + c} \right)\left( {{b^2}\; + {c^2}\;-bc} \right)}\\{ = \left( {b + c} \right)\left[ {{{\left( {b + c} \right)}^2}\;-3bc} \right]}\\{ = {{\left( {b + c} \right)}^3}\;-3bc\left( {b + c} \right)}\\{ \Rightarrow {a^3}\; + {b^3}\; + {c^3}\;-3abc = {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc}\\{ \Leftrightarrow {a^3}\; + {b^3}\; + {c^3}\;-3abc = {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc\left( {b + c} \right)-3abc}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}} \right)-\left[ {3bc\left( {b + c} \right) + 3abc} \right]}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}} \right)-3bc\left( {a + b + c} \right)}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}\;-3bc} \right)}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-ab\; - ac + {b^2}\; + 2bc + {c^2}\;-3bc} \right)}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\; + {b^2}\; + {c^2}\;-ab-ac-bc} \right)}\end{array}\)

Do đó nếu \({a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = 0\) thì \(a + b + c\; = 0\) hoặc \({a^2}\; + {b^2}\; + {c^2}\;-ab-ac-bc = 0\)

Mà \({a^2}\; + {b^2}\; + {c^2}\;-ab-ac-bc = .\left[ {{{\left( {a-b} \right)}^2}\; + {{\left( {a-c} \right)}^2}\; + {{\left( {b-c} \right)}^2}} \right]\)

Nếu \({\left( {a-b} \right)^2}\; + {\left( {a-c} \right)^2}\; + {\left( {b-c} \right)^2}\; = 0 \Leftrightarrow \;\left\{ \begin{array}{l}a - b = 0\\b - c = 0\\a - c = 0\end{array} \right. \Rightarrow a = b = c\)

Vậy \({a^3}\; + \left( {{b^3}\; + {c^3}} \right) = 3abc\) thì \(a = b = c\) hoặc \(a + b + c = 0\).