Trắc nghiệm Bài 7: Lập phương của một tổng hay một hiệu Toán 8 Kết nối tri thức

\({\left( {A + B} \right)^3}\; = {A^3}\; + 3{A^2}B + 3A{B^2}\; + {B^3}\); \({\left( {A\; - B} \right)^3}\; = {A^3}\; - 3{A^2}B + 3A{B^2}\; - {B^3}\)

\(\begin{array}{l}A + \frac{3}{4}{x^2} - \frac{3}{2}x + 1 = A + 3.{\left( { - \frac{1}{2}x} \right)^2}.1 + 3.\left( { - \frac{1}{2}x} \right){.1^2} + {1^3} = {\left( { - \frac{1}{2}x} \right)^3} + 3.{\left( { - \frac{1}{2}x} \right)^2}.1 + 3.\left( { - \frac{1}{2}x} \right){.1^2} + {1^3} = {\left( { - \frac{x}{2} + 1} \right)^3}\\ \Rightarrow A = {\left( { - \frac{1}{2}x} \right)^3} =- \frac{{{x^3}}}{8};\,B =- \frac{1}{2}x =- \frac{x}{2}\end{array}\)

\({23^3} - {9.23^2} + 27.23 - 27 \\= {23^3} - {3.23^2}.3 + {3.23.3^2} - {3^3} \\= {\left( {23 - 3} \right)^3} \\= {20^3} = 8000\)

\(8-{{ 36}}x + {{ 54}}{x^2}\;-{{ 27}}{x^3} = {2^3} - {3.2^2}.\left( {3x} \right) + 3.2.{\left( {3x} \right)^2} - {\left( {3x} \right)^3} = {\left( {2 - 3x} \right)^3}\)

\({x^3}\;-6{x^2}y + 12x{y^2}\;-8{y^3}\; = {x^3}\;-3.{x^2}.\left( {2y} \right) + 3.x.{\left( {2y} \right)^2} - {\left( {2y} \right)^3} = {\left( {x - 2y} \right)^3}\)

Thay \(x = 2021\) và \(y = 1010\) vào biểu thức trên ta có\({\left( {2021 - 2.1010} \right)^3} = {1^3} = 1\)

\(\begin{array}{l}{x^3}\;-12{x^2}\; + 48x-64 = 0 \Leftrightarrow {x^3}\;-{{ 3}}.{x^2}.4 + 3.x{.4^2} - {4^3} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\left( {x - 4} \right)^3} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x - 4 = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x = 4\end{array}\)

\({\left( {A - B} \right)^3}\; = {A^3}\; - 3{A^2}B + 3A{B^2}\; - {B^3}\)và phép nhân đa thức với đơn thức rồi tìm đưa về bài toán tìm \(x\) đã biết.

\(\begin{array}{l}H = \left( {x + 5} \right)({x^2}\;-5x + 25)-{\left( {2x + 1} \right)^3}\; + 7{\left( {x-1} \right)^3}\;-3x\left( { - 11x + 5} \right)\\\,\,\,\,\,\,\,\, = {x^3} - 5{x^2} + 25x + 5{x^2} - 25x + 125 - \left( {8{x^3} + 12{x^2} + 6x + 1} \right) + 7\left( {{x^3} - 3{x^2} + 3x - 1} \right) + 33{x^2} - 15x\\\,\,\,\,\,\,\,\, = {x^3} + 125 - 8{x^3} - 12{x^2} - 6x - 1 + 7{x^3} - 21{x^2} + 21x - 7 + 33{x^2} - 15x\\\,\,\,\,\,\,\,\, = \left( {{x^3} - 8{x^3} + 7{x^3}} \right) + \left( { - 12{x^2} - 21{x^2} + 33{x^2}} \right) + \left( {{5^3} - 1 - 7} \right)\\\,\,\,\,\,\,\,\, = 117\end{array}\)

Vậy \(H\) là một số lẻ.

\(\begin{array}{l}M = {\left( {x + 2y} \right)^3} - 6{\left( {x + 2y} \right)^2} + 12\left( {x + 2y} \right) - 8\\\,\,\,\,\,\,\, = {\left( {x + 2y} \right)^3} - 3.{\left( {x + 2y} \right)^2}.2 + 3.\left( {x + 2y} \right){.2^2} - {2^3}\\\,\,\,\,\,\,\, = {\left( {x + 2y - 2} \right)^3}\end{array}\)

Thay \(x = 20;\,y = 1\) vào biểu thức \(M\) ta có \(M = {\left( {20 + 2.1 - 2} \right)^3} = {20^3} = 8000\).

\({\left( {A - B} \right)^3}\; = {A^3}\; - 3{A^2}B + 3A{B^2}\; - {B^3}\) và phép nhân hai đa thức rồi thu gọn đa thức.

\(\begin{array}{l}P = {\left( {4x + 1} \right)^3}\;-\left( {4x + 3} \right)\left( {16{x^2}\; + 3} \right)\\\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {4x} \right)}^3}\; + 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2}\; + {1^3}\;-(64{x^3}\; + 12x + 48{x^2}\; + 9)}\\\begin{array}{l} = 64{x^3}\; + 48{x^2}\; + 12x + 1-64{x^3}\;-12x-48{x^2}\;-9\\ = - 8\end{array}\end{array}\end{array}\)

\(\begin{array}{l}Q = {\left( {x-2} \right)^3}\;-x\left( {x + 1} \right)\left( {x-1} \right) + 6x\left( {x-3} \right) + 5x\\\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {x^3}\;-3.{x^2}.2 + 3x{{.2}^2}\;-{2^3}\;-x\left( {{x^2}\;-1} \right) + 6{x^2}\;-18x + 5x}\\\begin{array}{l} = {x^3}\;-6{x^2}\; + 12x-8-{x^3}\; + x + 6{x^2}\;-18x + 5x\\ = - 8\end{array}\end{array}\end{array}\)

\( \Rightarrow P = Q\)

\({\left( {A - B} \right)^2}\; = {A^2}\; - 2AB + {B^2}\)

\(\begin{array}{l}P = 8{x^3}\;-12{x^2}y + 6x{y^2}\;-{y^3}\; + 12{x^2}\;-12xy + 3{y^2}\; + 6x-3y + 11\\\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {2x-y} \right)}^3}\; + 3{{\left( {2x-y} \right)}^2}\; + 3\left( {2x-y} \right) + 1 + 10}\\{\; = {{\left( {2x-y + 1} \right)}^3}\; + 10}\end{array}\end{array}\)

Ta có

\(\begin{array}{l}Q = {\left( {2x-{\rm{ 1}}} \right)^3}\;-{\rm{ 8}}x\left( {x + 1} \right)\left( {x-1} \right) + {\rm{ 2}}x\left( {6x - 5} \right)\\\,\,\,\,\,\,\, = 8{x^3} - 12{x^2} + 6x - 1 - 8x\left( {{x^2} - 1} \right) + 12{x^2} - 10x\\\,\,\,\,\,\,\, = 8{x^3} - 12{x^2} + 6x - 1 - 8{x^3} + 8x + 12{x^2} - 10x\\\,\,\,\,\,\,\, = 4x - 1\\ \Rightarrow a = 4;\,\,b = 1\end{array}\)

\(\begin{array}{l}\;\,\,\,\,\,\,{(x + 1)^3} - {(x - 1)^3} - 6{(x - 1)^2} = 20\\ \Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - \left( {{x^3} - 3{x^2} + 3x - 1} \right) - 6\left( {{x^2} - 2x + 1} \right) = - 10\\ \Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} + 3{x^2} - 3x + 1 - 6{x^2} + 12x - 6 = - 10\\ \Leftrightarrow 12x - 4 = 20\\ \Leftrightarrow 12x = 20 + 4\\ \Leftrightarrow 12x = 24\\ \Leftrightarrow x = 2\end{array}\)

\( \Rightarrow a = 2\). Vậy ước của \(2\) là \(2\).

Ta có

\(\begin{array}{l}\;P = {\left( {4x + 1} \right)^3}\;-\left( {4x + 3} \right)(16{x^2}\; + 3)\\\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {4x} \right)}^3}\; + 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2}\; + {1^3}\;-\left( {64{x^3}\; + 12x + 48{x^2}\; + 9} \right)}\\\begin{array}{l} = 64{x^3}\; + 48{x^2}\; + 12x + 1-64{x^3}\;-12x-48{x^2}\;-9\\ = - 8\end{array}\end{array}\\Q = {\left( {x-2} \right)^3}\;-x\left( {x + 1} \right)\left( {x-1} \right) + 6x\left( {x-3} \right) + 5x\\\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {x^3}\;-3.{x^2}.2 + 3x{{.2}^2}\;-{2^3}\;-x\left( {{x^2}\;-1} \right) + 6{x^2}\;-18x + 5x}\\\begin{array}{l} = {x^3}\;-6{x^2}\; + 12x-8-{x^3}\; + x + 6{x^2}\;-18x + 5x\\ = - 8\end{array}\end{array}\\ \Rightarrow P = Q\end{array}\)

Ta có

\(\begin{array}{l}\;A = 8{x^3}\;-12{x^2}y + 6x{y^2}\;-{y^3}\; + 12{x^2}\;-12xy + 3{y^2}\; + 6x-3y + 11\\\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{ = {{\left( {2x} \right)}^3}\;-3.{{\left( {2x} \right)}^2}.y + 3.2x.y + {y^3}\; + 3\left( {4{x^2}\;-4xy + {y^2}} \right) + 3\left( {2x-y} \right) + 11}\\{\; = {{\left( {2x-y} \right)}^3}\; + 3{{\left( {2x-y} \right)}^2}\; + 3\left( {2x-y} \right) + 1 + 10}\\{\; = {{\left( {2x-y + 1} \right)}^3}\; + 10}\end{array}\end{array}\)

Thay \(\;2x-y = 9\) vào biểu thức \(\;A\) ta có \(\;A = {\left( {9 + 1} \right)^3} + 10 = 1010\)

Ta có

\({(a - b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3} = {a^3} - {b^3} - 3ab(a - b)\)

Suy ra \( {a^3} - {b^3} = {(a - b)^3} + 3ab(a - b)\)

hay \(Q = {(a - b)^3} + 3ab(a - b)\)

Thay \(a + b = 5\) và \(ab = - 3\) vào Q ta có

\(\begin{array}{c}Q = {(a - b)^3} + 3ab(a - b)\\ = {4^3} + 3.( - 3).4\\ = 64 - 36\\ = 28\end{array}\)

\(\begin{array}{c}{(a + b + c)^3} = {{\rm{[}}(a + b) + c{\rm{]}}^3}\\ = {(a + b)^3} + 3{(a + b)^2}c + 3(a + b){c^2} + {c^3}\\ = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} + 3{(a + b)^2}c + 3(a + b){c^2} + {c^3}\\ = {a^3} + {b^3} + {c^3} + 3ab(a + b) + 3{(a + b)^2}c + 3(a + b){c^2}\\ = {a^3} + {b^3} + {c^3} + 3(a + b)\left[ {ab + (a + b)c + {c^2}} \right]\\ = {a^3} + {b^3} + {c^3} + 3(a + b)(ab + ac + bc + {c^2})\\ = {a^3} + {b^3} + {c^3} + 3(a + b)\left[ {a(b + c) + c(b + c)} \right]\\ = {a^3} + {b^3} + {c^3} + 3(a + b)(b + c)(c + a)\end{array}\)

Vậy \({(a + b + c)^3}\) = \({a^3} + {b^3} + {c^3} + 3(a + b)(b + c)(c + a)\)

\(\begin{array}{l}\;{(a + b)^3}\; = {a^3}\; + 3{a^2}b + 3a{b^2}\; + {b^3}\; = {a^3}\; + {b^3}\; + 3ab\left( {a + b} \right)\\ \Rightarrow {a^3}\; + {b^3}\; = {\left( {a + b} \right)^3}\;-3ab\left( {a + b} \right)\end{array}\)

Ta có:

\(\begin{array}{c}\;B = {a^3}\; + {b^3}\; + {c^3}\;-3abc\;\\ = {(a + b)^3} - 3ab(a + b) + {c^3} - 3abc\\ = {(a + b)^3} + {c^3} - 3ab(a + b + c)\end{array}\)

Tương tự, ta có \({(a + b + c)^3} - 3(a + b)c(a + b + c)\)

\( \Rightarrow B = {(a + b + c)^3} - 3(a + b)c(a + b + c) - 3ab(a + b + c)\)

Mà \(\;a + b + c = 0\) nên \(\;B = 0 - 3(a + b)c.0 - 3ab.0 = 0\)