Question

Tìm x :

a) 4x² - 9 = 0

b) 2x² + 0,36 = 1

c) \(\frac{5}{12}.\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)

d) 3x² + 7 = -4

Answer 1

a, 4x² - 9 = 0 => (2x)² = 9 => 2x = 3 hoặc 2x = -3 => x = 3/2 hoặc x = -3/2

b, 2x² + 0,36 = 1 => 2x² = 0,64 => x² = 0,32 = 8/25 => \(\orbr{\begin{cases}x=\sqrt{\frac{8}{25}}\\x=-\sqrt{\frac{8}{25}}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2\sqrt{2}}{5}\\x=\frac{-2\sqrt{2}}{5}\end{cases}}\)

c, \(\frac{5}{12}.\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)

\(\Rightarrow\frac{5}{12}.\sqrt{x}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)

\(\Rightarrow\sqrt{x}=\frac{1}{2}\div\frac{5}{12}\)

\(\Rightarrow\sqrt{x}=\frac{6}{5}\)

\(\Rightarrow x=\left(\frac{6}{5}\right)^2=\frac{36}{25}\)

d, 3x² + 7 = -4 => 3x² = -4 - 7 => 3x² = -11 => x² = -11/3 (vô lý) => x ∈ Ø