Bài 1:
a: \(\Leftrightarrow\left(2x+1\right)^2=64\)
=>2x+1=8 hoặc 2x+1=-8
=>x=7/2 hoặc x=-9/2
b: \(\Leftrightarrow6+3n-2n+10=10\)
=>6+n=0
hay n=-6
c: \(\Leftrightarrow8x-4-6x^2+3x=0\)
=>\(-6x^2+11x-4=0\)
hay \(x\in\left\{\dfrac{4}{3};\dfrac{1}{2}\right\}\)