Bài 42 trang 15 SBT toán 7 tập 1

Đề bài

Tìm \(x ∈\mathbb Q\), biết rằng:

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

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

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

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

Lời giải chi tiết

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

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

\(\Rightarrow x = \displaystyle {1 \over 2}\)  

Vậy \(x = \displaystyle {1 \over 2}\)

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

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

\(\Rightarrow x - 2 = 1 \) hoặc \(x - 2 = - 1 \)

\(\Rightarrow x  = 1+2 \) hoặc \(x  = - 1 +2\)

\(\Rightarrow x = 3 \) hoặc \(x = 1\)

Vậy \(x = 3 \) hoặc \(x = 1\)

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

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

\(\Rightarrow 2{\rm{x}} - 1 =  - 2\)

\( \Rightarrow 2x =  - 2 + 1\)

\( \Rightarrow 2x =  - 1\)

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

Vậy \(x = \displaystyle  - {1 \over 2}\)

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

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

\( \Rightarrow x + \dfrac{1}{2} = \dfrac{1}{4}\) hoặc \(x + \dfrac{1}{2} = - \dfrac{1}{4}\)

+) \(x + \dfrac{1}{2} = \dfrac{1}{4}\)\( \Rightarrow x = \dfrac{1}{4} - \dfrac{1}{2} =- \dfrac{1}{4}\)

+) \(x + \dfrac{1}{2} = - \dfrac{1}{4}\) \( \Rightarrow x = - \dfrac{1}{4} - \dfrac{1}{2}= - \dfrac{3}{4}\)

Vậy \(x=- \dfrac{1}{4}\) hoặc \(x=- \dfrac{3}{4}\)

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