Question #223090

Solve the inequality

x>√(x+2)

IxI / Ix+1I -1 > 1


1
Expert's answer
2021-09-01T11:28:03-0400

1.


x>x+2x>\sqrt{x+2}

We know that

x+20x+20\begin{matrix} x+2\geq0 \\ \\ \sqrt{x+2}\geq0 \end{matrix}

Then we have x>0x>0


x2>x+2x^2>x+2

x2x2>0x^2-x-2>0

(x+1)(x2)>0(x+1)(x-2)>0

x<1 or x>2x<-1\ or \ x>2

Since x>0,x>0, then we take x>2.x>2.


Answer: x>2.x>2.


x(2,).x\in(2, \infin).


2.


xx+11>1\dfrac{|x|}{|x+1|-1} > 1

x+110=>x0,x1|x+1|-1\not=0=>x\not=0, x\not=1


{x+11>0x>x+11\begin{cases} |x+1|-1>0 \\ |x|>|x+1|-1 \end{cases}

{x+1>1x+1>x+1\begin{cases} |x+1|>1 \\ |x|+1>|x+1| \end{cases}

x1,x2x\leq-1, x\not=-2

{x1>1x+1>x1\begin{cases} -x-1>1 \\ -x+1>-x-1 \end{cases}

{x<22>0\begin{cases} x<-2 \\ 2>0 \end{cases}

x(,2)x\in(-\infin, -2)


1x<0-1\leq x<0

{x+1>1x+1>x+1\begin{cases} x+1>1 \\ -x+1>x+1 \end{cases}

{x>02x<0\begin{cases} x>0 \\ 2x<0 \end{cases}

No solution



x>0x>0

{x+1>1x+1>x+1\begin{cases} x+1>1 \\ x+1>x+1 \end{cases}

{x>01>1\begin{cases} x>0 \\ 1>1 \end{cases}

No solution


Answer: x<2x<-2


x(,2)x\in(-\infin, -2)


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