Câu 1:
a: \(\Leftrightarrow6x^2-2x=6x^2-13\)
=>-2x=-13
hay x=13/2
b: \(\Leftrightarrow2x-2x-1=x-6x\)
=>-5x=-1
hay x=1/5
c: \(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=x^2+3\)
\(\Leftrightarrow x^2+3=x^2+2x+1-x^2+2x-1=4x\)
\(\Leftrightarrow x^2-4x+3=0\)
=>(x-1)(x-3)=0
=>x=3(nhận) hoặc x=1(loại)
Bài 4:
\(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+16x^2=0\)
\(\Leftrightarrow\left(x^2+x+4\right)^2+2.4x\left(x^2+x+4\right)+\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x^2+x+4+4x\right)^2=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2+4x+x+4\right)^2=0\)
\(\Leftrightarrow\left[x\left(x+4\right)+x+4\right]^2=0\)
\(\Leftrightarrow\left(x+1\right)^2.\left(x+4\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\) hay \(\left(x+4\right)^2=0\)
\(\Leftrightarrow x=-1\) hay \(x=-4\)
-Vậy \(S=\left\{-1;-4\right\}\)
