Toán tìm x sao cho 2√2 -1 = √x / √x -1 giúp minh vs 09/07/2021 By Emery tìm x sao cho 2√2 -1 = √x / √x -1 giúp minh vs
Đáp án: `2\sqrt{2-1}=\sqrtx/(\sqrtx-1)` `đk:x>=0,x ne 1` `pt<=>2=\sqrtx/(\sqrtx-1)` `<=>2\sqrtx-2=sqrtx` `<=>sqrtx=2` `<=>x=4(tm)` Vậy `x=4` thì `2\sqrt{2-1}=\sqrtx/(\sqrtx-1)`. Trả lời
ĐK : `x \ne 1 , x \ge 0` `2\sqrt{2-1} = \sqrt{x}/(\sqrt{x}-1)` `⇔ 2\sqrt{2-1}(\sqrt{x} – 1) = \sqrt{x}/(\sqrt{x}-1) (\sqrt{x}-1)` `⇔ 2(\sqrt{x}-1) = \sqrt{x}` `⇔ 2\sqrt{x}-2=\sqrt{x}` `⇔ 2\sqrt{x} = \sqrt{x} + 2` `⇔ \sqrt{x} = 2` `⇔ x = 4(TM)` Vậy `S ={4}` Trả lời
Đáp án:
`2\sqrt{2-1}=\sqrtx/(\sqrtx-1)`
`đk:x>=0,x ne 1`
`pt<=>2=\sqrtx/(\sqrtx-1)`
`<=>2\sqrtx-2=sqrtx`
`<=>sqrtx=2`
`<=>x=4(tm)`
Vậy `x=4` thì `2\sqrt{2-1}=\sqrtx/(\sqrtx-1)`.
ĐK : `x \ne 1 , x \ge 0`
`2\sqrt{2-1} = \sqrt{x}/(\sqrt{x}-1)`
`⇔ 2\sqrt{2-1}(\sqrt{x} – 1) = \sqrt{x}/(\sqrt{x}-1) (\sqrt{x}-1)`
`⇔ 2(\sqrt{x}-1) = \sqrt{x}`
`⇔ 2\sqrt{x}-2=\sqrt{x}`
`⇔ 2\sqrt{x} = \sqrt{x} + 2`
`⇔ \sqrt{x} = 2`
`⇔ x = 4(TM)`
Vậy `S ={4}`