Tìm x, biết:
a)\({\left( {0,5} \right)^2} + 2.x = {\left( {0,7} \right)^2}\)
b)\(x - \left( {\dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7}} \right) = \dfrac{1}{7} - \dfrac{1}{3}\)
Áp dụng quy tắc chuyển vế để tìm x.
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a)
\(\begin{array}{l}{\left( {0,5} \right)^2} + 2.x = {\left( {0,7} \right)^2}\\ \Rightarrow 2x = {\left( {0,7} \right)^2} - {\left( {0,5} \right)^2}\\ \Rightarrow 2x = 0,49 - 0,25\\ \Rightarrow 2x = 0,24\\ \Rightarrow x = 0,24:2\\ \Rightarrow x = 0,12\end{array}\)
Vậy \(x=0,12\)
b)
\(\begin{array}{l}x - \left( {\dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7}} \right) = \dfrac{1}{7} - \dfrac{1}{3}\\ \Rightarrow x = \dfrac{1}{7} - \dfrac{1}{3} + \left( {\dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7}} \right)\\ \Rightarrow x = \dfrac{1}{7} - \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7}\\ \Rightarrow x = \dfrac{1}{5}\end{array}\)
Vậy \(x = \dfrac{1}{5}\)