Question

Answer 1

14:

a: =>căn x-1+1=2

=>căn x-1=1

=>x-1=1

=>x=2

b: =>\(\left(\dfrac{2x}{1-\sqrt{2x+1}}\right)^2=2x+6\)

=>\(\left(\dfrac{2x+1-1}{\sqrt{2x+1}-1}\right)^2=2x+6\)

=>(căn 2x+1+1)^2=2x+6

=>2x+1+1+2*căn 2x+1=2x+6

=>2*căn(2x+1)=2x+6-2x-2=4

=.căn 2x+1=2

=>2x+1=4

=>2x=3

=>x=3/2

c: =>\(\left(\dfrac{x}{\sqrt{x+1}+1}\right)^2=x-4\)

=>\(\left(\sqrt{x+1}-1\right)^2=x-4\)

=>\(x+1+1-2\sqrt{x+1}=x-4\)

=>2căn (x+1)=x+2-x+4=6

=>x+1=9

=>x=8