\(6x-6=5x^2-5\)
\(\Leftrightarrow5x^2-6x-5+6=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
\(6x-6=5x^2-5\Leftrightarrow6\left(x-1\right)=5\left(x^2-1\right)\Leftrightarrow6\left(x-1\right)=5\left(x+1\right)\left(x-1\right)\Leftrightarrow6=5x+5\Leftrightarrow5x=6-5\Leftrightarrow5x=1\Leftrightarrow x=\dfrac{1}{5}\)
6x−6=5x2−56x−6=5x2−5
⇔5x2−6x−5+6=0⇔5x2−6x−5+6=0
⇔5x2−6x+1=0⇔5x2−6x+1=0
⇔5x2−5x−x+1=0⇔5x2−5x−x+1=0
⇔5x(x−1)−(x−1)=0⇔5x(x−1)−(x−1)=0
⇔(5x−1)(x−1)=0⇔(5x−1)(x−1)=0
⇔[5x−1=0x−1=0⇔[5x−1=0x−1=0
⇔⎡⎣x=15x=1
nếu hay tick nha
6x - 6 = 5x2 - 5
\(\Leftrightarrow\) 6(x - 1) = 5(x2 - 1)
\(\Leftrightarrow\) 6(x - 1) - 5(x - 1)(x + 1) = 0
\(\Leftrightarrow\) (x - 1)[6 - 5(x + 1)] = 0
\(\Leftrightarrow\) (x - 1)(6 - 5x - 5) = 0
\(\Leftrightarrow\) (x - 1)(1 - 5x) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-1=0\\1-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt!
Ta có: \(6x-6=5x^2-5\)
\(\Leftrightarrow5x^2-5-6x+6=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{1}{5}\right\}\)
