a) \(\left(x+2\right)^2-\left(x-1\right)\left(x+1\right)=-1\)
\(\Leftrightarrow x^2+4x+4-x^2+1+1=0\)
\(\Leftrightarrow4x+6=0\)
\(\Leftrightarrow4x=-6\Leftrightarrow x=-\dfrac{3}{2}\)
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b) sửa đề: \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x\right)^2-4^2-\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-4-1\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{5}{3}\end{matrix}\right.\)
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