Đáp án:
`a)` `S={2;-1}`
`b)` `S={15}`
`c)` `S={2}`
`d)` `S={-28}`
Giải thích các bước giải:
Câu `2` :
`a)` `sqrt[(2x-1)^2]=3`
`<=>|2x-1|=3`
`<=>[(2x-1=3),(2x-1=-3):}`
`<=>[(2x=4),(2x=-2):}`
`<=>[(x=2),(x=-1):}`
Vậy `S={2;-1}`
`b)` `sqrt(16x+16)-sqrt(9x+9)+sqrt(4x+4)+sqrt(x+1)=16` (ĐK : `x>=-1`)
`<=>sqrt[16(x+1)]-sqrt[9(x+1)]+sqrt[4(x+1)]+sqrt(x+1)=16`
`<=>4sqrt(x+1)-3sqrt(x+1)+2sqrt(x+1)+sqrt(x+1)=16`
`<=>4sqrt(x+1)=16`
`<=>sqrt(x+1)=4`
`<=>x+1=16`
`<=>x=15`(tm)
Vậy `S={15}`
`c)` `sqrt(x-1)+sqrt(4x-4)-sqrt(25x-25)+2=0` (ĐK : `x>=1`)
`<=>sqrt(x-1)+sqrt[4(x-1)]-sqrt[25(x-1)]=-2`
`<=>sqrt(x-1)+2sqrt(x-1)-5sqrt(x-1)=-2`
`<=>-2sqrt(x-1)=-2`
`<=>sqrt(x-1)=1`
`<=>x-1=1`
`<=>x=2` (tm)
Vậy `S={2}`
`d)` `sqrt(4-5x)=12`
ĐK : `4-5x>=0<=>-5x>=-4<=>x<=4/5`
`<=>4-5x=144`
`<=>5x=-140`
`<=>x=-28` (tm)
Vậy `S={-28}`.
