Question

Answer 1

b: =>(x-4)(2x-3-1)=0

=>(x-4)(2x-4)=0

=>x=2 hoặc x=4

d: =>(x-2020)(x-3)(x+3)=0

hay \(x\in\left\{2020;-3;3\right\}\)

Answer 2

\(\left(b\right):\left(2x-3\right)\left(x-4\right)-x+4=0\\ < =>\left(2x-3\right)\left(x-4\right)-\left(x-4\right)=0\\ < =>\left(x-4\right)\left(2x-3-1\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\2x-4=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.=>S=\left\{4;2\right\}\\ \left(d\right):\left(x^2-8\right)\left(x-2020\right)-x+2020=0\\ < =>\left(x^2-8\right)\left(x-2020\right)-\left(x-2020\right)=0\\ < =>\left(x-2020\right)\left(x^2-8-1\right)=0\\ =>\left[{}\begin{matrix}x-2020=0\\x^2-9=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2020\\x=\pm3\end{matrix}\right.=>S=\left\{2020;3;-3\right\}\)

 

Answer 3

\(\left(f\right):x^3-5x^2+x-5=0\\ < =>x^2\left(x-5\right)+\left(x-5\right)=0\\ < =>\left(x-5\right)\left(x^2+1\right)=0\\ =>\left[{}\begin{matrix}x-5=0\\x^2+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x^2=-1< 0\left(KTM\right)\end{matrix}\right.=>S=\left\{5\right\}\\ \left(h\right):x\left(x^2+1\right)-2x\left(x-2\right)-2\left(x^2+1\right)=0\\ < =>\left(x^2+1\right)\left(x-2\right)-2x\left(x-2\right)=0\\ < =>\left(x-2\right)\left(x^2+1-2x\right)=0\\ =>\left[{}\begin{matrix}x-2=0\\\left(x-1\right)^2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.=>S=\left\{2;1\right\}\)