Tìm x, biết:
a) \({\rm{}}{x \over { - 15}} = {{ - 60} \over x}\)
b) \({{ - 2} \over x} = {{ - x} \over {{8 \over {25}}}}\)
Hướng dẫn giải
\(\begin{array}{l}
+ )\,\,\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow ad = bc\\
+ )\,\,{A^2} = {B^2}\,\left( {B > 0} \right) \Rightarrow \left[ \begin{array}{l}
A = B\\
A = - B
\end{array} \right.
\end{array}\)
Lời giải chi tiết
a) \(\displaystyle {x \over { - 15}} = {{ - 60} \over x}\)
\(\Rightarrow x.x = \left( { - 15} \right).\left( { - 60} \right) \)
\(\Rightarrow {x^2} = 900\)
\( \Rightarrow {x^2} = {30^2}\)
\(\Rightarrow x = 30\) hoặc \(x = -30\)
Vậy \(x = 30\) hoặc \(x = -30\)
b) \(\displaystyle {{ - 2} \over x} = {{ - x} \over {\displaystyle {8 \over {25}}}}\)
\(\displaystyle \Rightarrow - 2.{8 \over {25}} = x.\left( { - x} \right) \)
\(\displaystyle \Rightarrow - {{16} \over {25}} =- {x^2}\)
\( \displaystyle \Rightarrow {x^2} = {{16} \over {25}}\)
\( \Rightarrow {x^2} = {\left( {\dfrac{4}{5}} \right)^2}\)
\(\displaystyle \Rightarrow x = {4 \over 5}\) hoặc \(\displaystyle x = - {4 \over 5}\)
Vậy \(\displaystyle x = {4 \over 5}\) hoặc \(\displaystyle x = - {4 \over 5}\)
-- Mod Toán 7