a) Đề sai.
b) \(\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0-1\\\sqrt{x}=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=9\end{matrix}\right.\)
Vậy \(x=9.\)
c) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
Chúc bạn học tốt!
a) \(\left|2x+1\right|+\left|x+8\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|2x+1\right|\ge0\\\left|x+8\right|\ge0\end{matrix}\right.\forall x.\)
Do đó \(4x\ge0\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(2x+1\right)+\left(x+8\right)=4x\)
\(\Rightarrow\left(2x+x\right)+\left(1+8\right)=4x\)
\(\Rightarrow3x+9=4x\)
\(\Rightarrow4x-3x=9\)
\(\Rightarrow1x=9\)
\(\Rightarrow x=9:1\)
\(\Rightarrow x=9\)
Vậy \(x=9.\)
Chúc bạn học tốt!
