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Câu hỏi: Tìm \(x ∈\mathbb Q\), biết rằng:
\({\rm{a}}) {\left( {x - \displaystyle {1 \over 2}} \right)^2} = 0\)
\(b) {\left( {x - 2} \right)^2} = 1\)
\(c) {\left( {2{\rm{x}} - 1} \right)^3} = - 8\)
\({\rm{d}}) {\left( {x + \displaystyle {1 \over 2}} \right)^2} = \displaystyle {1 \over {16}}\)
\({\rm{a}}) {\left( {x - \displaystyle {1 \over 2}} \right)^2} = 0\)
\(b) {\left( {x - 2} \right)^2} = 1\)
\(c) {\left( {2{\rm{x}} - 1} \right)^3} = - 8\)
\({\rm{d}}) {\left( {x + \displaystyle {1 \over 2}} \right)^2} = \displaystyle {1 \over {16}}\)
Phương pháp giải
\(\begin{array}{l}
+ ) {A^2} = {B^2} \Rightarrow \left[ \begin{array}{l}
A = B\\
A = - B
\end{array} \right.\\
+ ) {A^3} = {B^3} \Rightarrow A = B\\
+ ) {A^2} = 0 \Rightarrow A = 0
\end{array}\)
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 \(x=- \dfrac{3}{4}\)
\(\begin{array}{l}
+ ) {A^2} = {B^2} \Rightarrow \left[ \begin{array}{l}
A = B\\
A = - B
\end{array} \right.\\
+ ) {A^3} = {B^3} \Rightarrow A = B\\
+ ) {A^2} = 0 \Rightarrow A = 0
\end{array}\)
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 \(x=- \dfrac{3}{4}\)