Question

1. \(2\left(2x^2+4x+3\right)=\left(5x+4\right)\sqrt{x^2+3}\)(làm liên hợp thì khó cm biểu thức dài dằng dặc kia vô nghiệm)

2. Tìm GTLN của \(A=\frac{21x+\sqrt{5x+1}+9}{7x+3}\)

Answer 1

\(\left(1\right)< =>-3\left(x-1\right)\left(x+1\right)\left(3x^2-8x-4\right)=0=>\orbr{\begin{cases}x=1\\x=\frac{4-2\sqrt{7}}{3};\frac{4+2\sqrt{7}}{3}\end{cases}.}\)
 

Answer 2

\(A=3+\frac{\sqrt{5x+1}}{7x+3}=3+\frac{\sqrt{5x+1}}{\frac{7}{5}\left(5x+1\right)+\frac{8}{5}}=3+\frac{1}{\frac{1}{5}\left(7\sqrt{5x+1}+\frac{8}{\sqrt{5x+1}}\right)}\le3+\frac{1}{\frac{1}{5}2\sqrt{7.8}}=...\)