\(ĐKXĐ:\frac{-1}{4}\le x\le3\)
\(PT\Leftrightarrow3x+14-6\sqrt{4x+1}-2\sqrt{3-x}=0\)
\(\Leftrightarrow\left(4x+1\right)-2.3\sqrt{4x+1}+9+\left(3-x\right)-3\sqrt{3-x}+1=0\)
\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2=0\)(1)
Mà \(\left(\sqrt{4x+1}-3\right)^2\ge0\forall x\);\(\left(\sqrt{3-x}-1\right)^2\ge0\forall x\)
\(\Rightarrow\)(1) xảy ra khi \(\hept{\begin{cases}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x+1=9\\3-x=1\end{cases}}\Rightarrow x=2\left(tmđk\right)\)
Vậy nghiệm duy nhất của phương trình là 2
\(6\sqrt[]{4x+1}+2\sqrt[]{3-x}=3x+14\)
Ấn nhầm sorry Chờ làm lại
\(6\sqrt{4x+1}+2\sqrt{3-x}=3x+14\)
=> \(3x+14-6\sqrt{4x+1}-2\sqrt{3-x}=0\)
=>\(\left(4x+1-2\sqrt{4x+1}.3+9\right)+\left(3-x-2\sqrt{3-x}+1\right)=0\)
=>\(\left(\sqrt{4x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2=0\)
=>\(\hept{\begin{cases}\sqrt{4x+1}-3=0\\\sqrt{3-x}-1=0\end{cases}}\)tự tìm x nhé
