Question

Giúp mình với

Answer 1

b: \(\Leftrightarrow10x^2=6\)

\(\Leftrightarrow x^2=\dfrac{3}{5}=\dfrac{15}{25}\)

hay \(x\in\left\{\dfrac{\sqrt{15}}{5};-\dfrac{\sqrt{15}}{5}\right\}\)

c: \(\Leftrightarrow x^2-9-3x+5=0\)

\(\Leftrightarrow x^2-3x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)

=>x=4 hoặc x=-1

Answer 2

\(b,10x^2-6=0.\\ \Leftrightarrow10x^2=6.\\ \Leftrightarrow x^2=\dfrac{3}{5}.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}.\\x=\dfrac{-3}{5}.\end{matrix}\right.\)

\(c,\left(x-3\right)\left(x+3\right)=3x-5.\\ x^2-9-3x+5=0.\\ \Leftrightarrow x^2-3x-4=0.\\ \Leftrightarrow\left(x-4\right)\left(x+1\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-1.\end{matrix}\right.\)

\(d,x^2-2\sqrt{3}x-6=0.\\ \Delta'=3+6=9>0.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\sqrt{3}+3.\\x_2=\sqrt{3}-3.\end{matrix}\right.\)