\(4.\left(x+1\right)^2-9.\left(x-1\right)^2=0\)
\(\Leftrightarrow4.\left(x^2+2x+1\right)-9.\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow4x^2+8x+4-9x^2+18x-9=0\)
\(\Leftrightarrow\left(4x^2+8x+4\right)-\left(9x^2-18x+9\right)=0\)
\(\Leftrightarrow\left(2x+2\right)^2-\left(3x-3\right)^2=0\)
\(\Leftrightarrow\left[2x+2-\left(3x-3\right)\right].\left[2x+2+\left(3x-3\right)\right]=0\)
\(\Leftrightarrow\left(2x+2-3x+3\right).\left(2x+2+3x-3\right)=0\)
\(\Leftrightarrow\left(5-x\right).\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-x=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{5;\frac{1}{5}\right\}.\)
Chúc bạn học tốt!
\(4\left(x+1\right)^2-9\left(x-1\right)^2=0\)
\(\Leftrightarrow4\left(x^2+2x+1\right)-9\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow4x^2+8x+4-9x^2-18x-9=0\)
\(\Leftrightarrow-5x^2-10x-5=0\)
\(\Leftrightarrow-5\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow-5\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy S = {1}
