`a)\sqrt{2x-5}=\sqrt{x-1}` `ĐK: x >= 5/2`
`<=>2x-5=x-1`
`<=>x=4` (t/m)
Vậy `S={4}`
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`b)x^2-2=x-2\sqrt{x+1}` `ĐK: x >= -1`
`<=>x^2+2x+1=3x+3-2\sqrt{x+1}`
`<=>(x+1)^2=3(x+1)-2\sqrt{x+1}`
Đặt `\sqrt{x+1}=t` `(t >= 0)`
Ptr có dạng:
`t^4=3t^2-2t`
`<=>t^4-3t^2+2t=0`
`<=>[(t=0),(t=1),(t=2):}` (t/m)
`=>[(\sqrt{x+1}=0),(\sqrt{x+1}=1),(\sqrt{x+1}=2):}`
`<=>[(x+1=0),(x+1=1),(x+1=4):}`
`<=>[(x=-1),(x=0),(x=3):}` (t/m)
Vậy `S={-1;0;3}`
`(a):\sqrt{2x-5}=\sqrt{x-1}\ (ĐK:x\ge 5/2)`
`<=>2x-5=x-1`
`<=>2x-x=5-1`
`<=>x=4(TMDK)`
Vậy `S={4}`
`(b):x^{2}-2=x-2\sqrt{x+1}\ (ĐK:x\ge -1)`
`<=>x^{2}=x+1-2.\sqrt{x+1}.1+1^{2}`
`<=>x^{2}=(\sqrt{x+1}-1)^{2}`
`<=>|x|=|\sqrt{x+1}-1|`
`=>x=\sqrt{x+1}-1\ (1)` hoặc `x=-\sqrt{x+1}+1\ (2)`
`(1)<=>x+1=\sqrt{x+1}`
`<=>x^{2}+2x+1=x+1` `(ĐK:x\ge -1)`
`<=>x^{2}+x=0`
`<=>x(x+1)=0`
`=>x=0` hoặc `x=-1` (TMDK)
`(2)<=>\sqrt{x+1}=1-x`
`<=>x+1=x^{2}-2x+1` `(ĐK:1-x\ge0;x\ge -1<=>-1\le x\le 1)`
`<=>x^{2}-3x=0`
`<=>x(x-3)=0`
`=>x=0\ (TMDK)` hoặc `x=3\ (KTMDK)`
Vậy `S={0;-1}`
b) Đề bài
--> \(x^2-2-x=2\sqrt{x+1}\)
-->\(x^4-2x^3-3x^2+4x+4=4\left(x+1\right)\)
-->\(x^4-2x^2+4x+4=4x+4\)
-->\(x^4-2x^3-3x^2+4=4\)
-->\(x^4-2x^3-3x^2=0\)
-->\(x^2\left(x^2-2x-3\right)=0\)
--> x=0 , -1
Lời giải:
a. ĐKXĐ: $x\geq \frac{5}{2}$
PT $\Leftrightarrow 2x-5=x-1$ (bình phương 2 vế)
$\Leftrightarrow 2x-5-(x-1)=0$
$\Leftrightarrow x-4=0$
$\Leftrightarrow x=4$ (tm)
b.
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow x^2-x-2+2\sqrt{x+1}=0$
$\Leftrightarrow (x-2)(x+1)+2\sqrt{x+1}=0$
$\Leftrightarrow \sqrt{x+1}[(x-2)\sqrt{x+1}+2]=0$
$\Leftrightarrow \sqrt{x+1}=0$ hoặc $(x-2)\sqrt{x+1}+2=0$
Nếu $\sqrt{x+1}=0\Leftrightarrow x=-1$ (tm)
Nếu $(x-2)\sqrt{x+1}+2=0$
$\Leftrightarrow (2-x)\sqrt{x+1}=2$
$\Rightarrow x<2$
Bình phương 2 vế:
$(2-x)^2(x+1)=4$
$\Leftrightarrow (x^2-4x+4)(x+1)=4$
$\Leftrightarrow x^3-3x^2=0$
$\Leftrightarrow x^2(x-3)=0$
$\Leftrightarrow x=0$ hoặc $x=3$
Do $x\geq -1$ và $x<2$ nên $x=0$
Vậy $x\in\left\{-1;0\right\}$
