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`B5=(15sqrtx-11)/(x+2sqrtx-3)+(3sqrtx-2)/(1-sqrtx)-(2sqrtx+3)/(3+sqrtx)(x>=0;xne1)`
`=(15sqrtx-11)/((sqrtx-1)(sqrtx+3))+(2-3sqrtx)/(sqrtx-1)-(2sqrtx+3)/(sqrtx+3)`
`=(15sqrtx-11+(2-3sqrtx)(sqrtx+3)+(1-sqrtx)(2sqrtx+3))/((sqrtx-1)(sqrtx+3))`
`=(15sqrtx-11+2sqrtx-3x+6-9sqrtx+2sqrtx-2x+3-3sqrtx)/((sqrtx-1)(sqrtx+3))`
`=(-5x+7sqrtx-2)/((sqrtx-1)(sqrtx+3))`
`=(-(sqrtx-1)(5sqrtx-2))/((sqrtx-1)(sqrtx+3))`
`=(2-5sqrtx)/(sqrtx+3)`
`B5=1/2`
`,=>(2-5sqrtx)/(sqrtx+3)=1/2`
`=>2(2-5sqrtx)=sqrtx+3`
`<=>4-10sqrtx=sqrtx+3`
`<=>1=11sqrtx`
`<=>sqrtx=1/11`
`<=>x=1/121(tm)`
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`a)` Rút Gọn `B5`
`B5=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)+(3\sqrt{x}-2)/(1-\sqrt{x})-(2\sqrt{x}+3)/(3+\sqrt{x})` với `x>=0;x\ne1`
`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3\sqrt{x}-2)/(\sqrt{x}-1)-(2\sqrt{x}+3)/(\sqrt{x}+3)`
`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-((3\sqrt{x}-2)(\sqrt{x}+3))/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}+3)(\sqrt{x}-1))`
`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+9\sqrt{x}-2\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}+3)(\sqrt{x}-1))`
`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+7\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}+3)(\sqrt{x}-1))`
`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+7\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-(2x-2\sqrt{x}+3\sqrt{x}-3)/((\sqrt{x}+3)(\sqrt{x}-1))`
`=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)-(3x+7\sqrt{x}-6)/((\sqrt{x}-1)(\sqrt{x}+3))-(2x+\sqrt{x}-3)/((\sqrt{x}+3)(\sqrt{x}-1))`
`=(15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3)/((\sqrt{x}-1)(\sqrt{x}+3))`
`=(7\sqrt{x}-5x-2)/((\sqrt{x}-1)(\sqrt{x}+3))`
`=(-(\sqrt{x}-1)(5\sqrt{x}-2))/((\sqrt{x}-1)(\sqrt{x}+3))`
`=(2-5\sqrt{x})/(\sqrt{x}+3)`
Vậy `B5=(2-5\sqrt{x})/(\sqrt{x}+3)` với `x>=0;x\ne1`
`b)` `B5=1/2`
`=>(2-5\sqrt{x})/(\sqrt{x}+3)=1/2`
`<=>2.(2-5\sqrt{x})=\sqrt{x}+3`
`<=>4-10\sqrt{x}=\sqrt{x}+3`
`<=>-10\sqrt{x}-\sqrt{x}=-4+3`
`<=>-11\sqrt{x}=-1`
`<=>\sqrt{x}=1/11`
`=>x=1/121`
Kết Hợp Với ĐKXĐ ta thấy `x=1/121` thoả mãn
Vậy `x=1/121` khi `B5=1/2`
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