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Question Number 92605 by som(math1967) last updated on 08/May/20

$$If\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=1$$$$findx$$

Commented by behi83417@gmail.com last updated on 08/May/20

$$\frac{(1+\mathrm{x})(1-\sqrt{1+\mathrm{x}})}{-\mathrm{x}}+\frac{(1-\mathrm{x})(1+\sqrt{1-\mathrm{x}})}{\mathrm{x}}=1$$$$\Rightarrow (1+\mathrm{x})\sqrt{1+\mathrm{x}}-(1+\mathrm{x})+(1-\mathrm{x})\sqrt{1-\mathrm{x}}+(1-\mathrm{x})=\mathrm{x}$$$$\Rightarrow (1+\mathrm{x})\sqrt{1+\mathrm{x}}+(1-\mathrm{x})\sqrt{1-\mathrm{x}}=3\mathrm{x}$$$$(-1<\mathrm{x}<1,\mathrm{x}\ne 0,\mathrm{let}:\mathrm{x}=\cos 2\mathrm{t})\Rightarrow$$$${2\mathrm{c}}^2.\sqrt{2}\mathrm{c}+{2\mathrm{s}}^2.\sqrt{2}\mathrm{s}=3({\mathrm{c}}^2-{\mathrm{s}}^2)$$$$\Rightarrow 2\sqrt{2}({\mathrm{c}}^3+{\mathrm{s}}^3)=3({\mathrm{c}}^2-{\mathrm{s}}^2)$$$$\Rightarrow 2\sqrt{2}(\mathrm{c}+\mathrm{s})({\mathrm{c}}^2+{\mathrm{s}}^2-\mathrm{cs})=3(\mathrm{c}+\mathrm{s})(\mathrm{c}-\mathrm{s})$$$$1.\mathrm{c}+\mathrm{s}=0\Rightarrow \sqrt{2}\cos (\mathrm{t}-\frac{\pi }{4})=0\Rightarrow \mathrm{t}=\frac{\pi }{4}(\equiv \mathrm{x}=0:\mathrm{not}\mathrm{ok})$$$$\Rightarrow 2\sqrt{2}(1-\mathrm{cs})=3(\mathrm{c}-\mathrm{s})$$$$\Rightarrow 8(1-2\mathrm{cs}+{\mathrm{c}}^2{\mathrm{s}}^2)=9({\mathrm{c}}^2+{\mathrm{s}}^2-2\mathrm{cs})$$$$\Rightarrow {8\mathrm{c}}^2{\mathrm{s}}^2+2\mathrm{cs}-1=0$$$$\Rightarrow \mathrm{cs}=\frac{-1\pm \sqrt{1+8}}{8}=\frac{1}{4},-\frac{1}{2}$$$$1.\mathrm{cs}=\frac{1}{4}\Rightarrow \sin 2\mathrm{t}=\frac{1}{2}\Rightarrow \mathbf{x}=\cos 2\mathrm{t}=\frac{\sqrt{3}}{2}$$$$2.\mathrm{cs}=-\frac{1}{2}\Rightarrow \sin 2\mathrm{t}=-1\Rightarrow \mathrm{x}=\cos 2\mathrm{t}=0\left[\mathrm{not}\mathrm{ok}\right]$$

Commented by Tony Lin last updated on 08/May/20

$$let\sqrt{1+x}=a,\sqrt{1-x}=b,a^2+b^2=2$$$$\frac{a^2}{1+a}+\frac{b^2}{1-b}=1$$$$a^2-a^{2}b+b^2+{ab}^2=1+a-b-ab$$$$1-ab(a-b)-(a-b)+ab=0$$$$(1+ab)[1-(a-b)]=0$$$$a-b=1orab=-1\left(false\right)$$$$\sqrt{1+x}-\sqrt{1-x}=1$$$$1+x={(1+\sqrt{1-x})}^2$$$$2x-1=2\sqrt{1-x}$$$$4x^2-4x+1=4-4x$$$$x^2=\frac{3}{4}$$$$x=\pm \frac{\sqrt{3}}{2}$$$$but\frac{1-\frac{\sqrt{3}}{2}}{1+\sqrt{1-\frac{\sqrt{3}}{2}}}+\frac{1+\frac{\sqrt{3}}{2}}{1-\sqrt{1+\frac{\sqrt{3}}{2}}}<0(-5)$$$$\therefore x=\frac{\sqrt{3}}{2}$$

Commented by som(math1967) last updated on 08/May/20

$$Thankssir$$

Commented by som(math1967) last updated on 08/May/20

$$Thankssir$$

Answered by MJS last updated on 08/May/20

$$-1⩽x⩽1\wedge x\ne 0$$$$\frac{(1+x)(1-\sqrt{1+x})}{x}+\frac{(1-x)(1+\sqrt{1-x})}{x}=1$$$${(1+x)}^{3/2}+{(1-x)}^{3/2}=3x$$$$0<x⩽1$$$$x={\sin }^{2}t$$$${\cos }^{3}t+{(1+{\sin }^{2}t)}^{3/2}={3\sin }^{2}t$$$${(1+{\sin }^{2}t)}^3={({3\sin }^{2}t-{\cos }^{3}t)}^2$$$$\sin t=s\wedge \cos t=c\wedge s=\sqrt{1-c^2}$$$$c^6+3c^5+\frac{3}{2}{c}^4-3c^3-3c^2+\frac{1}{2}=0$$$$c=z-1$$$$z^6-3z^5+\frac{3}{2}{z}^4+z^3=0$$$$z^3(z-2)(z^2-z-\frac{1}{2})=0$$$$(c-1){(c+1)}^3(c^2+c-\frac{1}{2})=0$$$$\cos t=\pm 1\vee \cos t=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}$$$$\Rightarrow \mathrm{only}\mathrm{solution}\mathrm{is}$$$$\cos t=-\frac{1}{2}+\frac{\sqrt{3}}{2}\Rightarrow {\sin }^{2}t=x=\frac{\sqrt{3}}{2}$$

Commented by som(math1967) last updated on 08/May/20

$$Thankyousir$$

Commented by MJS last updated on 08/May/20

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{also}\mathrm{possible}\mathrm{to}\mathrm{square}\mathrm{it}:$$$${(1-x)}^{3/2}+{(1+x)}^{3/2}=3x$$$$6x^2+2+2{(1-x)}^{3/2}{(1+x)}^{3/2}=9x^2$$$$2{(1-x)}^{3/2}{(1+x)}^{3/2}=3x^2-2$$$$4{(1-x)}^3{(1+x)}^3={(3x^2-2)}^2$$$$4x^6-3x^4=0$$$$x^4(4x^2-3)=0$$$$0<x⩽1\Rightarrow x=\frac{\sqrt{3}}{2}$$

Commented by som(math1967) last updated on 08/May/20

$$Nicesirthanksagain$$

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