`a)sqrt{x^2+4}=x-2`
`đk:x-2>=0<=>x>=2`
BP 2 vế ta có:
`x^2+4=x^2-4x+4`
`<=>4x=0`
`<=>x=0(l)`
Vậy pt vô nghiệm
`b)sqrt{x^2-10x+25}=3-19x`
`đk:3-19x>=0<=>x<=3/19`
`pt<=>\sqrt{(x-5)^2}=3-19x`
`<=>|x-5|=3-19x`
`+)x-5=3-19x`
`<=>20x=8<=>x=2/5(tm)`
`+)x-5=19x-3`
`<=>18x=-2`
`<=>x=-1/9(tm)`
Vậy `S={2/5,-1/9}`
`c)sqrt{x^2-9}+sqrt{x^2-6x+9}=0`
`đk:x^2-9>=0<=>x^2>=9`
Vì `sqrt{x^2-9}>=0`
`sqrt{x^2-6x+9}>=0`
`=>sqrt{x^2-9}=sqrt{x^2-6x+9}=0`
`<=>sqrt{(x-3)(x+3)}=sqrt{(x-3)^2}=0`
`<=>x=3`
Vậy `S={3]`
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