\(a,\dfrac{x+5}{7}=0\\ \Rightarrow x+5=0\\ \Rightarrow x=0-5\\ \Rightarrow x=-5\\ b,\dfrac{5x+1}{6}=0\\ \Rightarrow5x+1=0\\ \Rightarrow5x=0-1\\ \Rightarrow5x=-1\\ \Rightarrow x=-\dfrac{1}{5}\)
\(c,\dfrac{4x^2+3}{8}=0\\ \Rightarrow4x^2+3=0\\ \Rightarrow4x^2=0-3\\ \Rightarrow4x^2=-3\\ \Rightarrow x^2=-\dfrac{3}{4}\\ \Rightarrow x\in\varnothing\)
\(d,\dfrac{2x\left(7x^2-7\right)}{4}=0\\ \Rightarrow2x\left(7x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\7x^2-7=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\7x^2=7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
\(a,\) Để \(\dfrac{x+5}{7}=0\) thì : \(x+5=0\Leftrightarrow x=-5\)
\(b,\) Để \(\dfrac{5x+1}{6}=0\) thì : \(5x+1=0\Leftrightarrow5x=-1\Leftrightarrow x=-\dfrac{1}{5}\)
\(c,\) Để \(\dfrac{4x^2+3}{8}=0\) thì \(4x^2+3=0\Leftrightarrow4x^2=-3\Leftrightarrow x^2=-\dfrac{3}{4}\)
Vậy không có x thỏa mãn để biểu thức \(\dfrac{4x^2+3}{8}=0\)
\(d,\) Để \(\dfrac{2x\left(7x^2-7\right)}{4}=0\) thì \(2x\left(7x^2-7\right)=0\)
\(\Leftrightarrow2x\left[7\left(x^2-1\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
