Question

1) Giải phương trình (2đ)

a) \(\dfrac{x-1}{x-2}-\dfrac{1}{x+2}=\dfrac{x^2-4}{x^2-4}\)                            b) | 3x+2 | = x-1

c) 5x ( x+1 ) - 15 ( x+1 ) = 0                              d) 6x + 5 = 7x - 3

Answer 1

 

\(b,x-1\ge0\\ x\ge1\\ TH1\left|3x+2\right|=3x+2\\ 3x+2=x-1\\ 3x-x=-1-2\\ 2x=-3\\ x=-\dfrac{3}{2}\left(kothoaman\right)\\ TH2\left|3x+2\right|=-3x-2\\ -3x-2=x-1\\ -3x-x=-1+2\\ -4x=1\\ x=-\dfrac{1}{4}\left(kothoaman\right)=>\varnothing\)

\(c,\left(x+1\right)\left(5x-15\right)=0\\ \left[{}\begin{matrix}x+1=0\\5x-15=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ d,6x+5-7x+3=0\\ -x+8=0\\ -x=-8\\ x=8\)