Question

Answer 1

`1)sqrt[(2x-1)^2]=3`

`<=> |2x-1| =3`

`<=>[(2x-1=3),(2x-1=-3):}`

`<=>[(x=2),(x=-1):}`

`2)5/3sqrt(15x) -sqrt(15x)-2 =1/3sqrt(15x)(đk:x>=0)`

`<=> (5/3-1-1/3)sqrt(15x) = 2`

`<=> 1/3sqrt(15x)=2`

`<=> sqrt{15x} =6`

`<=> 15x =36`

`=>x = 12/5(t//m)`

3)`sqrt(x-6) (x^2-1)=0(đk:x>=6)`

`<=> [(sqrt(x-6)=0),(x^2-1)=0:}`

`<=> [(x-6=0),(x^2=1):}`

`<=> [(x=6(t//m)),(x=+-1(loại)):}`

4)`2sqrt(x-3)=1/3sqrt(4x-12)(đk:x>=3)`

`<=>2sqrt(x-3) = 2/3sqrt(x-3)`

`<=> 4/3sqrt(x-3)=0`

`<=> x-3=0`

`<=> x=3(t//m)`

`5)sqrt(4(x+1))=sqrt(8)(đk:x>=-1)`

`<=> 4(x+1)=8`

`<=> x+1=2`

`<=> x=1(t//m)`

Answer 2

5: =>4(x+1)=8

=>x+1=2

=>x=1

4: =>2*căn x-3=2/3*căn x-3

=>x-3=0

=>x=3

5: =>x-6=0 hoặc x^2-1=0

=>x=1; x=-1;x=6