Question

a) 9x²=25

b) x³-\(\dfrac{1}{4}\)x=0

c) 9x²-36=0

Answer 1

\(a)9.x^2=25\)

\(\Leftrightarrow x^2=\dfrac{25}{9}\)

\(\Leftrightarrow x^2=\left(\pm\dfrac{5}{9}\right)^2\)

\(\Leftrightarrow x=\pm\dfrac{5}{9}\)

\(\Leftrightarrow x\in\left\{\dfrac{5}{9};-\dfrac{5}{9}\right\}\)

Answer 2

b, \(x^3-\dfrac{1}{4}x=0\)

⇔ \(x\left(x^2-\dfrac{1}{4}\right)=0\)

⇔ \(x\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)=0\)

⇔\(\left[{}\begin{matrix}x=0\\x+\dfrac{1}{2}\\x-\dfrac{1}{2}\end{matrix}\right.=0\) ⇔ \(\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy xϵ \(\left\{0,\pm\dfrac{1}{2}\right\}\)

Answer 3

a, 9x² = 25

⇔ x² = \(\dfrac{25}{9}\)

⇔ x² = \(\dfrac{5}{3}\)

⇔ x² = \(\left(\dfrac{5}{3}\right)^2\)

⇔ x ² = \(\pm\dfrac{5}{3}\)

Vậy x = \(\pm\dfrac{5}{3}\)

Answer 4

\(b)x^3-\dfrac{1}{4}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Answer 5

\(c)9x^2-36=0\)

\(\Leftrightarrow9x^2=36\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=\left(\pm2\right)^2\)

\(\Leftrightarrow x=\pm2\)

\(\Leftrightarrow x\in\left\{2;-2\right\}\)

Answer 6

a, \(9x^2=25\Rightarrow9x^2=\left(\sqrt{25}\right)^2\)

\(\Rightarrow9x=\pm\sqrt{25}\Rightarrow9x=\pm5\)

\(\Rightarrow x=\pm\dfrac{5}{9}\)

b, \(x^3-\dfrac{1}{4}x=0\Rightarrow x^2\left(x-\dfrac{1}{4}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

c, \(9x^2-36=0\Rightarrow9x^2=36\)

\(\Rightarrow9x^2=6^2\Rightarrow9x=6\)

\(\Rightarrow x=\dfrac{2}{3}\)

Answer 7

OH! Phần c mình làm sai r nhé :V

Answer 8

c, \(9x^2-36=0\)

⇔ \(3\left(3x^2-12\right)=0\)

⇔ 3x² -12=0

⇔ 3x² = 12

⇔x² = 4

⇔ x² = \(\pm2\)

Vậy x = \(\pm2\)