Giải bài 45 trang 20 – SGK Toán lớp 8 tập 1
Tìm \(x,\) biết:
a) \(2 - 25x^2 = 0;\)
b) \(x^2 - x + \dfrac{1}{4}\)
Gợi ý:
a) \(2-25x^2=(\sqrt 2)^2+(5x)^2\)
b) \(x^2-x+\dfrac 1 4 =x^2-2.\dfrac 1 2 .x +\dfrac 1 4\)
a) \(2 - 25x^2 = 0\)
\(\Leftrightarrow (\sqrt{2})^2 - (5x)^2 = 0\)
\(\Leftrightarrow (\sqrt{2} - 5x)(\sqrt{2} + 5x ) = 0\)
\(\Rightarrow \left[\begin{array}{l}\sqrt{2} - 5x = 0 \\ \sqrt{2} + 5x = 0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{l} 5x = \sqrt{2} \\ 5x = -\sqrt{2} \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\sqrt{2}}{5} \\ x = -\dfrac{\sqrt{2}}{5} \end{array} \right.\)
b) \(x^2 - x + \dfrac{1}{4} = 0\)
\(\Leftrightarrow x^2 - 2.x. \dfrac{1}{2} + \left(\dfrac{1}{2}\right)^2 = 0\)
\(\Leftrightarrow \left(x - \dfrac{1}{2}\right)^2 = 0\)
\(\Leftrightarrow x - \dfrac{1}{2} = 0\)
\(\Leftrightarrow x = \dfrac{1}{2}\)
Lưu ý: \(A.B = 0 \Leftrightarrow \left[\begin{array}{l} A = 0 \\ B = 0\end{array} \right.\)