\(a,\left(2x+3\right)^2=36\\ \left[{}\begin{matrix}2x+3=6\\2x+3=-6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\\ b,\left(3x+6\right)^2=\left(x-10\right)^2\\ 3x+6=x-10\\ 3x-x+6+10=0\\ 2x+16=0\\ 2x=-16\\ x=-8\\ c,\left[{}\begin{matrix}x+1=0\\2x-4=0\end{matrix}\right.=>x=-1;x=2\\ d,\left(2x+3\right)^3=64\\ 2x+3=4\\ 2x=1\\ x=\dfrac{1}{2}\\ b,x+6=2x-1\\ x-2x=-1-6\\ -x=-7\\ x=7\)
a: =>(2x+3+6)(2x+3-6)=0
=>(2x+9)(2x-3)=0
=>x=3/2 hoặc x=-9/2
b: =>(3x+6+x-10)(3x+6-x+10)=0
=>(4x-4)(2x+16)=0
=>x=-8 hoặc x=1
c: =>(x+1)=0 hoặc 2x-4=0
=>x=2 hoặc x=-1
d: =>(2x+3+8)(2x+3-8)=0
=>(2x+11)(2x-5)=0
=>x=5/2 hoặc x=-11/2
