\(a,x\left(x-3\right)=4\\ \Rightarrow x^2-3x-4=0\\ \Rightarrow\left(x^2-4x\right)+\left(x-4\right)=0\\ \Rightarrow x\left(x-4\right)+\left(x-4\right)=0\\ \Rightarrow\left(x-4\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\\ b,x\left(2x+1\right)=3\\ \Rightarrow2x^2+x-3=0\\ \Rightarrow\left(2x^2-2x\right)+\left(3x-3\right)=0\\ \Rightarrow2x\left(x-1\right)+3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,x\left(x+1\right)=6\\ \Rightarrow x^2+x-6=0\\ \Rightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\\ \Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\\ \Rightarrow\left(x-2\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\\ d,x\left(3x-5\right)=8\\ \Rightarrow3x^2-5x-8=0\\ \Rightarrow\left(3x^2+3x\right)-\left(8x+8\right)=0\\ \Rightarrow3x\left(x+1\right)-8\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(3x-8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
a,
\(x\left(x-3\right)=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Leftrightarrow x^2-4x+x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S\in\left\{4;-1\right\}\)
b,
\(x\left(2x+1\right)=3\)
\(\Leftrightarrow2x^2+2-3=0\)
\(\Leftrightarrow x=\sqrt{\left(0+3-2\right):2}=\dfrac{\sqrt{2}}{2}\)
Vậy pt có tập nghiệm \(S\in\left\{\dfrac{\sqrt{2}}{2}\right\}\)
c,
\(x\left(x+1\right)=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S\in\left\{2;-3\right\}\)
d,
\(x\left(3x-5\right)=8\)
\(\Leftrightarrow3x^2-5x-8=0\)
\(\Leftrightarrow3x^2-8x+3x-8=0\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(8x+8\right)=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\3x-8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S\in\left\{-1;\dfrac{8}{3}\right\}\)
