\(\frac{x-1}{7}=\frac{9}{x+1}\)
\(\Leftrightarrow\left(x-1\right).\left(x+1\right)=9.7\)
Tới đây VT ta áp dụng hằng đẳng thức số 3 .
\(x^2-1^2=63\)
\(x^2=63+1\)
\(x^2=64\)
\(\Rightarrow x=\pm\sqrt{63}\)
\(\frac{x-1}{7}=\frac{9}{x+1}\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=9.7\)
\(\Rightarrow x-1=7;x+1=9\Rightarrow x=8\)
\(\frac{x-1}{7}=\frac{9}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=9\cdot7\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-8\end{array}\right.\)
Apd dụng tc của dãy tỉ số bằng nhau ta có
\(\frac{x-1}{7}=\frac{9}{x+1}=\frac{x-1+9}{7+x+1}=\frac{x+8}{x+8}=1\)
\(\Rightarrow\begin{cases}x-1=7\\x+1=9\end{cases}\)
=> x=8 (TM)
\(\frac{x-1}{7}=\frac{9}{x+1}\Rightarrow\left(x-1\right)\left(x+1\right)=7\cdot9\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x^2=8^2\) hoặc \(\left(-8\right)^2\)
\(\Rightarrow x=\pm8\)
\(\Rightarrow\left(x+1\right)\left(x-1\right)=9.7=63\)
\(\Rightarrow x^2-x+x-1=63\Rightarrow x^2-1=63\Rightarrow x^2=64\)
\(\Rightarrow x=8ho\text{ặ}c-8\)
\(\frac{x-1}{7}=\frac{9}{x+1}\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=9\cdot7\)
\(x^2-1=63\)
\(x^2=63+1\)
\(x^2=64\)
\(\Rightarrow x=\pm8\)
Vậy \(x=\pm8\)