Câu 1:
\(\sqrt{2x+1}=3\)
\(\Rightarrow2x+1=9\)
\(\Rightarrow x=4\)
Vậy x = 4
Câu 2:
\(\left(2x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{2}\) hoặc \(x=-3\)
Câu 3: Sai đề
1) \(\sqrt[]{2x+1}=3\)
\(\Leftrightarrow2x+1=3^2\)
\(\Leftrightarrow2x=9-1\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
2)\(\left(2x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{2}\) hoặc \(x=-3\)
3) \(5x=2y=x^2+y\)
Ta có:
\(5x=2y\)
\(\Leftrightarrow5\div2\times x=y\)
\(\Leftrightarrow\dfrac{5}{2}x=y\)
\(x^2+y=2y\)
\(\Leftrightarrow x^2=y\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow x=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
