Question

giải pt
a)(3x-7)(2x^2-1)=0
b)(5+2x)(4x^4+3)=0

 

Answer 1

a.

\(\left(3x-7\right)\left(2x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\2x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x^2=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\pm\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

b.

\(\left(5+2x\right)\left(4x^4+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x^4=-\dfrac{3}{4}< 0\left(vô-nghiệm\right)\end{matrix}\right.\)

Vậy \(x=-\dfrac{5}{2}\)