Câu 42 trang 15 Sách Bài Tập (SBT) Toán 7 tập 1

Tìm x ∈ Q, biết rằng:

\({\rm{a}})\;{\left( {x - {1 \over 2}} \right)^2} = 0\)

\(b)\;{\left( {x - 2} \right)^2} = 1\)

\(c)\;{\left( {2{\rm{x}} - 1} \right)^3} =  - 8\)

\({\rm{d}})\;{\left( {x + {1 \over 2}} \right)^2} = {1 \over {16}}\)

Giải

\({\rm{a}})\;{\left( {x - {1 \over 2}} \right)^2} = 0 \)

\(\Rightarrow x - {1 \over 2} = 0 \)

\(\Rightarrow x = {1 \over 2}\) 

\(b)\;{\left( {x - 2} \right)^2} = 1 \)

\(\Rightarrow \left[ \matrix{
x - 2 = 1 \hfill \cr
x - 2 = - 1 \hfill \cr} \right.\)

\(\Rightarrow \left[ \matrix{
x = 3 \hfill \cr
x = 1 \hfill \cr} \right.\)

\(c)\;{\left( {2{\rm{x}} - 1} \right)^3} =  - 8\)

\(\Rightarrow {\left( {2{\rm{x}} - 1} \right)^3} = {\left( -2 \right)^3}\)

\(\Rightarrow 2{\rm{x}} - 1 =  - 2\)

\(\Rightarrow x =  - {1 \over 2}\)

\({\rm{d)}}\;{\left( {x + {1 \over 2}} \right)^2} = {1 \over {16}}\)

\(\Rightarrow {\left( {x + {1 \over 2}} \right)^2} = {\left( {{1 \over 4}} \right)^2} \)

\(\Rightarrow \left[ \matrix{
x + {1 \over 2} = {1 \over 4} \hfill \cr
x + {1 \over 2} = - {1 \over 4} \hfill \cr} \right. \)

\(\Rightarrow \left[ \matrix{
x = - {1 \over 4} \hfill \cr
x = - {3 \over 4} \hfill \cr} \right.\)

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