Cho hai biểu thức $\;P = {\left( {4x + 1} \right)^3}\;-\left( {4x + 3} \right)(16{x^2}\; + 3);\,\,Q = {\left( {x-2}…

Phương pháp giải:

Sử dụng hằng đẳng thức: ${\left( {A - B} \right)^3}\; = {A^3}\; - 3{A^2}B + 3A{B^2}\; - {B^3}$,${\left( {A - B} \right)^2}\; = {A^2}\; - 2AB + {B^2}$,${\left( {A + B} \right)^3}\; = {A^3}\; + 3{A^2}B + 3A{B^2}\; + {B^3}$ đưa về bài toán tìm $x$.

Lời giải chi tiết:

Ta có

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