HOC24
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Chủ đề / Chương
Bài học
`(x-2)\sqrt{x^2+5}=x^2-4`
`<=>(x-2)[\sqrt{x^2+5}-(x+2)]=0`
`<=>[(x=2),(\sqrt{x^2+5}=x+2):}`
`<=>[(x=2),({(x+2 >= 0),(x^2+5=x^2+4x+4):}):]`
`<=>[(x=2),({(x >= -2),(x=1/4):}):]`
`<=>[(x=2),(x=1/4):]`
`=> C`.
Gọi `(S)` có tâm `I(1;-2;3)`, bán kính `R=5`
`(P) nn (S)` là đường tròn `(O; R')`
Theo đề bài ta có: `R'=\sqrt{19}`
`=>d(I,(P))=\sqrt{5^2 - 19}=\sqrt{6}=[|a-2b+3+3|]/\sqrt{a^2+b^2+1}`
`=>6a^2+6b^2+6=a^2+4b^2+36-4ab+12a-24b`
`<=>5a^2+2b^2+4ab-12a+24b-30=0` `(1)`
Ta có `A in (P)=>-a+b-2+3=0<=>b=a-1` `(2)`
Từ `(1);(2)=>5a^2+2(a-1)^2+4a(a-1)-12a+24(a-1)-30=0`
`<=>11a^2+4a-52=0<=>[(a=2),(a=-26/11):}`
Mà `a in ZZ`
`=>a=2`
`=>b=1`
Vậy `a+b=3`.
`a)TXĐ:R\\{1;1/3}`
`y'=[-4(6x-4)]/[(3x^2-4x+1)^5]`
`b)TXĐ:R`
`y'=2x. 3^[x^2-1] ln 3-e^[-x+1]`
`c)TXĐ: (4;+oo)`
`y'=[2x-4]/[x^2-4x]+2/[(2x-1).ln 3]`
`d)TXĐ:(0;+oo)`
`y'=ln x+2/[(x+1)^2].2^[[x-1]/[x+1]].ln 2`
`e)TXĐ:(-oo;-1)uu(1;+oo)`
`y'=-7x^[-8]-[2x]/[x^2-1]`
`lim_{x->-oo}[3x^2-5x+2]/[x^4+2x^2-1]` `ĐK: x ne +-\sqrt{-1+\sqrt{2}}`
`=lim_{x->-oo}[3/[x^2]-5/[x^3]+2/[x^4]]/[1+2/[x^2]-1/[x^4]]`
`=0`
`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]` `ĐK: 4x^3+1 ne 0`
`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`
`=1/4`
`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]` `ĐK: x ne 0;x ne 1`
`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`
Câu `c` giống `b`.
`a)lim_{x->+oo}[x+1]/[x^2+x+1]`
`=lim_{x->+oo}[1/x+1/[x^2]]/[1+1/x+1/[x^2]]`
`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`
`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`
`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`
`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`
`=-3`
`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`
`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`
`=-5/3`
`a)lim_{x->3^[-]} [-4x]/[x-3]=+oo`
`b)lim_{x->3^[+]} [-4x]/[x-3]=-oo`
Ừm, mình nhầm.
`1)(a^[1/4]-b^[1/4])(a^[1/4]+b^[1/4])(a^[1/2]+b^[1/2])`
`=[(a^[1/4])^2-(b^[1/4])^2](a^[1/2]+b^[1/2])`
`=(a^[1/2]-b^[1/2])(a^[1/2]+b^[1/2])`
`=a-b`
`2)(a^[1/3]-b^[2/3])(a^[2/3]+a^[1/3]b^[2/3]+b^[4/3])`
`=(a^[1/3]-b^[2/3])[(a^[1/3])^2+a^[1/3]b^[2/3]+(b^[2/3])^2]`
`=(a^[1/3])^3-(b^[2/3])^3`
`=a-b^2`
`a)TXĐ: R`
`b)TXĐ: R\\{0}`
`c)TXĐ: R\\{1}`
`d)TXĐ: (-oo;-1)uu(1;+oo)`
`e)TXĐ: (-oo;-1/2)uu(1/2;+oo)`
`f)TXĐ: (-oo;-\sqrt{2})uu(\sqrt{2};+oo)`
`h)TXĐ: (-oo;0) uu(2;+oo)`
`k)TXĐ: R\\{1/2}`
`l)ĐK: {(x^2-1 > 0),(x-2 > 0),(x-1 ne 0):}`
`<=>{([(x > 1),(x < -1):}),(x > 2),(x ne 1):}`
`<=>x > 2`
`=>TXĐ: (2;+oo)`