\(\sqrt{3x+1}+\sqrt{2-x}=3\)
\(\Leftrightarrow\left(\sqrt{3x+1}+\sqrt{2-x}\right)^2=9\)
\(\Leftrightarrow3x+1+2.\sqrt{\left(3x+1\right)\left(2-x\right)}+2-x=0\)
\(\Leftrightarrow2x+3+2\sqrt{\left(3x+1\right)\left(2-x\right)}=9\)
\(\Leftrightarrow2x+2\sqrt{\left(3x+1\right)\left(2-x\right)}=6\)
\(\Leftrightarrow2\left(x+\sqrt{6x-3x^2+2-x}\right)=6\)
\(\Leftrightarrow x+\sqrt{5x-3x^2+2}\) = 3
\(\Leftrightarrow\sqrt{5x-3x^2+2}=3-x\)
\(\Leftrightarrow5x-3x^2+2=\left(3-x\right)^2\)
\(\Leftrightarrow5x-3x^2+2=9-6x+x^2\)
\(\Leftrightarrow5x-3x^2+2-9+6x-x^2=0\)
\(\Leftrightarrow11x-4x^2-7=0\)
\(\Leftrightarrow11x-11x^2+7x^2-7=0\)
\(\Leftrightarrow11x\left(1-x\right)-7\left(1-x^2\right)=0\)
\(\Leftrightarrow11x\left(1-x\right)-7\left(1+x\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(11x-7-7x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(4x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1-x=0\\4x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{4}\end{matrix}\right.\)
Sai thì thông cảm nha!!!hate
