Bài 9:
a) Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-4=9\)
\(\Leftrightarrow-12x-3=9\)
\(\Leftrightarrow-12x=12\)
hay x=-1
Vậy: S={-1}
b) Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x+41=1\)
\(\Leftrightarrow2x=-40\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow7x^2+8x+13-7x^2+63=0\)
\(\Leftrightarrow8x=-74\)
hay \(x=-\dfrac{37}{4}\)
Vậy: \(S=\left\{-\dfrac{37}{4}\right\}\)
d) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
Vậy: S={7}
e) Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\)
\(\Leftrightarrow12x-4=-19\)
\(\Leftrightarrow12x=-15\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
