Bài 1 :
a, \(\left(3x+1\right)\left(5-x\right)-7=0\)
=> \(15x+5-3x^2-x-7=0\)
=> \(-3x^2+14x-2=0\)
=> \(3x^2-14x+2=0\)
=> \(\left(x\sqrt{3}\right)^2-2.x.\sqrt{3}.\frac{7\sqrt{3}}{3}+\frac{49}{3}-\frac{43}{3}=0\)
=> \(\left(x\sqrt{3}-\frac{7\sqrt{3}}{3}\right)^2-\left(\sqrt{\frac{43}{3}}\right)^2=0\)
=> \(\left(x\sqrt{3}-\frac{7\sqrt{3}}{3}-\sqrt{\frac{43}{3}}\right)\left(x\sqrt{3}-\frac{7\sqrt{3}}{3}+\sqrt{\frac{43}{3}}\right)=0\)
=> \(\left[{}\begin{matrix}x\sqrt{3}-\frac{7\sqrt{3}}{3}-\sqrt{\frac{43}{3}}=0\\x\sqrt{3}-\frac{7\sqrt{3}}{3}+\sqrt{\frac{43}{3}}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\sqrt{3}=\frac{7\sqrt{3}}{3}+\sqrt{\frac{43}{3}}\\x\sqrt{3}=\frac{7\sqrt{3}}{3}-\sqrt{\frac{43}{3}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{\frac{7\sqrt{3}}{3}+\sqrt{\frac{43}{3}}}{\sqrt{3}}=\frac{7+\sqrt{43}}{3}\\x=\frac{\frac{7\sqrt{3}}{3}-\sqrt{\frac{43}{3}}}{\sqrt{3}}=\frac{7-\sqrt{43}}{3}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(\left[{}\begin{matrix}x=\frac{7+\sqrt{43}}{3}\\x=\frac{7-\sqrt{43}}{3}\end{matrix}\right.\)