Nội dung bài giảng
55. Tìm \(x\), biết:
a) \({x^3} - {1 \over 4}x = 0\);
b) \({(2x - 1)^2} - {(x + 3)^2} = 0\);
c) \({x^2}(x - 3) + 12 - 4x = 0\).
Bài giải:
a)
\(\eqalign{
& {x^3} - {1 \over 4}x = 0 \Rightarrow x\left( {{x^2} - {1 \over 4}} \right) = 0 \cr
& \Rightarrow x\left( {{x^2} - {{\left( {{1 \over 2}} \right)}^2}} \right) = 0 \cr
& \Rightarrow x\left( {x - {1 \over 2}} \right)\left( {x + {1 \over 2}} \right) = 0 \cr
& \Rightarrow \left[ \matrix{
x = 0 \hfill \cr
\left( {x - {1 \over 2}} \right) = 0 \Rightarrow x = {1 \over 2} \hfill \cr
\left( {x + {1 \over 2}} \right) = 0 \Rightarrow x = - {1 \over 2} \hfill \cr} \right. \cr} \)
Vậy \(x=0,x={1\over 2},x=-{1\over2}\)
b)
\(\eqalign{
& {(2x - 1)^2} - {(x + 3)^2} = 0 \cr
& \Rightarrow \left[ {(2x - 1) - (x + 3)} \right].\left[ {(2x - 1) + (x + 3)} \right] = 0 \cr
& \Rightarrow (2x - 1 - x - 3).(2x - 1 + x + 3) = 0 \cr
& \Rightarrow (x - 4).(3x + 2) = 0 \cr
& \Rightarrow \left[ \matrix{
x - 4 = 0 \hfill \cr
3x + 2 = 0 \hfill \cr} \right. \Rightarrow \left[ \matrix{
x = 4 \hfill \cr
x = - {2 \over 3} \hfill \cr} \right. \cr} \)
Vậy \(x=4,x=-{2\over 3}\)
c)
\(\eqalign{
& {x^2}(x - 3) + 12 - 4x = 0 \cr
& \Rightarrow {x^2}(x - 3) - 4(x - 3) = 0 \cr
& \Rightarrow (x - 3)({x^2} - 4) = 0 \cr
& \Rightarrow (x - 3)(x - 2)(x + 2) = 0 \cr
& \Rightarrow \left[ \matrix{
x = 3 \hfill \cr
x = 2 \hfill \cr
x = - 2 \hfill \cr} \right. \cr} \)
Vậy \( x=3,x=2,x=-2\)