Tìm x ∈ Q, biết rằng:
\({\rm{a}}){\left( {x - {1 \over 2}} \right)^2} = 0\)
\(b){\left( {x - 2} \right)^2} = 1\)
\(c){\left( {2{\rm{x}} - 1} \right)^3} = - 8\)
\({\rm{d}}){\left( {x + {1 \over 2}} \right)^2} = {1 \over {16}}\)
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\({\rm{a}}){\left( {x - {1 \over 2}} \right)^2} = 0 \Rightarrow x - {1 \over 2} = 0 \Rightarrow x = {1 \over 2}\)
\(b){\left( {x - 2} \right)^2} = 1 \Leftrightarrow \left[ \matrix{
x - 2 = 1 \hfill \cr
x - 2 = - 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 3 \hfill \cr
x = 1 \hfill \cr} \right.\)
\(c){\left( {2{\rm{x}} - 1} \right)^3} = - 8 \Rightarrow {\left( {2{\rm{x}} - 1} \right)^3} = {\left( -2 \right)^3}\)
\(\Rightarrow 2{\rm{x}} - 1 = - 2 \Rightarrow x = - {1 \over 2}\)
\({\rm{d)}}{\left( {x + {1 \over 2}} \right)^2} = {1 \over {16}} \Rightarrow {\left( {x + {1 \over 2}} \right)^2} = {\left( {{1 \over 4}} \right)^2} \)
\(\Leftrightarrow \left[ \matrix{
x + {1 \over 2} = {1 \over 4} \hfill \cr
x + {1 \over 2} = - {1 \over 4} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = - {1 \over 4} \hfill \cr
x = - {3 \over 4} \hfill \cr} \right.\)