Question

\(^{x^2}-\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\)

\(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\)

\(x^2-5=5\)

\(3x^2-1=14\)

Answer 1

\(x^2-\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\\ \Rightarrow x^2-\dfrac{1}{4}=\dfrac{3}{4}\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=-\dfrac{1}{3}\\\dfrac{1}{2}-x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)

\(x^2-5=5\\ \Rightarrow x^2=10\\ \Rightarrow x=\pm\sqrt{10}\)

\(3x^2-1=14\\ \Rightarrow3x^2=15\\ \Rightarrow x^2=5\\ \Rightarrow x=\pm\sqrt{5}\)

Answer 2

1. x² - \(\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\)

<=> x² - \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\)

<=> x² = \(\dfrac{3}{4}+\dfrac{1}{4}\)

<=> x² = 1

<=> x = \(\pm\)1

2. \(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\)

<=> \(\dfrac{1}{4}-x+x^2=\dfrac{1}{9}\)

<=> x² - x = \(\dfrac{1}{9}-\dfrac{1}{4}\)

<=> x² - x = \(\dfrac{-5}{36}\)

<=> x² - x - \(\dfrac{-5}{36}\) = 0

Đoạn này dài, mik giải ngoài rồi viết vào nha:

<=> x = \(\dfrac{5}{6}\)

3. x² - 5 = 5

<=> x² = 10

<=> x = \(\sqrt{10}\)

4. 3x² - 1 = 14

<=> 3x² = 15

<=> x² = 15 : 3

<=> x² = 5

<=> x = \(\sqrt{5}\)