Question

Bài 3

a/ 7x(x-5)-x+5=0

b/(x-3)^2-4(x+2)^2=0

Answer 1

\(a)7x(x-5)-x+5=0\\<=>7x(x-5)-(x-5)=0\\<=>(x-5)(7x-1)=0\\<=> \left[\begin{matrix} x-5=0\\ 7x-1=0\end{matrix}\right.\\<=> \left[\begin{matrix} x=5\\ x=1/7\end{matrix}\right.\\b)(x-3)^2-4(x+2)^2=0\\<=>(x-3)^2-[2x(x+2)]^2=0\\<=>(x-3)^2-(2x^2+4x)^2=0\\<=>(x-3-2x^2-4x)(x-3+2x^2+4x)=0\\<=>(-2x^2-5x-3)(2x^2+5x-3)=0\\<=>(-2x^2-2x-3x-3)(2x^2-x+6x-3)=0\\\)
\(<=>[-2x(x+1)-3(x+1)][x(2x-1)+3(2x-1)]\\<=>(-2x-3)(x+1)(2x-1)(x+3)=0\\TH1: -2x-3=0\\<=>x=3/2\\TH2:x+1=0\\<=>x=-1\\TH3:2x-1=0\\<=>x=1/2\\TH4:x+3=0\\<=>x=-3\)