6x2 - (2x - 3)(3x + 2) = 1
<=> 6x2 - (6x2 - 5x - 6) = 1
<=> 5x + 6 = 1
=> 5x = -5
=> x = -1
Vậy x = -1
b) (x + 1)3 - (x - 1)(x2 + x + 1) - 2 = 0
<=> x3 + 3x2 + 3x + 1 - (x3 - 1) - 2 = 0
<=> 3x2 + 3x = 0
<=> 3x(x + 1) = 0
<=> x(x + 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy \(x\in\left\{0;-1\right\}\)
\(a\text{)}6x^2\left(2x-3\right)\left(3x+2\right)=1\)
\(\Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\)
\(\Leftrightarrow5x+6=1\)
\(\Leftrightarrow5x=\left(-5\right)\)
\(\Rightarrow x=-1\)
\(b\text{)}\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-1\right)-2=0\)
\(\Leftrightarrow3x^2+3x=0\)
\(\Leftrightarrow3x\left(x+1=0\right)\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow x=0;x=\left(-1\right)\)
\(\Rightarrow x\in\left\{0;-1\right\}\)
\(\text{Hok tốt!}\)
\(\text{@Kaito Kid}\)
