1.a)=> (4x-5).\(\dfrac{1}{2x-1}\)
=>\(\dfrac{4x-5}{2x-1}\)
=>2x-5:(-1)
=>2x=5
=>x=5:2
Câu 2:
a: (x-1)(2y+1)=6
mà x,y là các số nguyên
nên \(\left(x-1;2y+1\right)\in\left\{\left(6;1\right);\left(2;3\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(7;0\right);\left(3;1\right)\right\}\)
b: Đặt x/3=y/4=k
=>x=3k; y=4k
Ta có: xy+1=13
=>12k2+1=13
=>k2=1
TH1: k=1
=>x=3; y=4
TH2: k=-1
=>x=-3; y=-4