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