Question

a,4x²-1=0            b,2x²+0,82=1

Answer 1

a)\(4x^2-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}}\)

Vậy\(x=\frac{1}{2};\frac{-1}{2}\)

Answer 2

b)\(2x^2+0,82=1\)

\(\Leftrightarrow2x^2=0,18\)

\(\Leftrightarrow x^2=0,09\)

\(\Leftrightarrow x^2=\frac{9}{100}\)

\(\Rightarrow x=\frac{3}{10};\frac{-3}{10}\)

Answer 3

a) 4x²  - 1 =0 

=> 4x²  = 1 

=>. x² = \(\frac{1}{4}\) 

=> x² = ( \(\frac{1}{2}\))^{2 => x = \(\frac{1}{2}\)}

^{vậy x =} ​\(\frac{1}{2}\)​​ 

b) 2x² +0,82 =1 => 2x² + \(\frac{41}{50}\) =1 => 2x²  = \(\frac{9}{50}\) 

                                                         => x² = \(\frac{9}{100}\) =>  x² = ( \(\frac{3}{10}\))²

                                                                       ^{=> x=\(\frac{3}{10}\)}  

                 Vậy x = \(\frac{3}{10}\)

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