9x² – 1 = (3x + 1)(4x + 1)
<=> (3x - 1)(3x + 1) = (3x + 1)(4x + 1)
<=> (3x−1)(3x+1)\3x+1 = (3x+1)(4x+1)\3x+1
=> 3x - 1 = 4x + 1
<=> 3x - 4x = 1 + 1
<=> -x = 2
<=> x = -2
Vậy x = -2
Ta có : \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
=> \(\left(3x-1\right)\left(3x+1\right)=\left(3x+1\right)\left(4x+1\right)\)
=> \(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
=> \(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
=> \(\left(3x+1\right)\left(x+2\right)=0\)
=> \(\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{-\frac{1}{3};-2\right\}\)
\(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow9x^2-1=12x^2+7x+1\)
\(\Leftrightarrow3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow3x+1=0ho\text{ặc}x+2=0\)
\(\Leftrightarrow x=-2ho\text{ặc}x=-\frac{1}{3}\)
Vậy \(x\in\left\{-2;-\frac{1}{3}\right\}\)
\(9x^2-1=\left(3x+1\right)\cdot\left(4x+1\right)\)
\(\Leftrightarrow\left(3x-1\right)\cdot\left(3x+1\right)-\left(3x+1\right)\cdot\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\cdot\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\cdot\left(-x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)
9x2 - 1 = (3x + 1)(4x + 1)
\(\Leftrightarrow\) (3x - 1)(3x + 1) = (3x + 1)(4x + 1)
\(\Leftrightarrow\) (3x - 1)(3x + 1) - (3x + 1)(4x + 1) = 0
\(\Leftrightarrow\) (3x + 1)(3x - 1 - 4x - 1) = 0
\(\Leftrightarrow\) (3x + 1)(-x - 2) = 0
\(\Leftrightarrow\) 3x + 1 = 0 hoặc -x - 2 = 0
\(\Leftrightarrow\) x = \(\frac{-1}{3}\) hoặc x = -2
Vậy S = {\(\frac{-1}{3}\); -2}
Chúc bạn học tốt!
