Question Number 108753 by bemath last updated on 19/Aug/20
$$\:\:\frac{\vdots\mathcal{B}{e}\mathcal{M}{ath}\vdots}{\bigtriangleup} \\ $$$$\:\frac{\sqrt{{x}}\:+\mathrm{1}}{{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}}\::\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:−{x}^{\mathrm{2}} }\:+\:{x}\:=\:?\: \\ $$
Commented by Rasheed.Sindhi last updated on 19/Aug/20
$${I}'{ve}\:{thought}\:{it}\:{as} \\ $$$$\:\frac{\sqrt{{x}}\:+\mathrm{1}}{{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}}\::\left(\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:−{x}^{\mathrm{2}} }\:+\:{x}\right) \\ $$$${because}\:{I}\:{think}\:':'{has}\:{low}\:{priority}\: \\ $$$${than}\:'+'\:\left({according}\:{to}\:{convention}\right) \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\:\frac{\sqrt{{x}}\:+\mathrm{1}}{{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}}\::\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:−{x}^{\mathrm{2}} }\:+\:{x}\:\: \\ $$$$\:\frac{\sqrt{{x}}\:+\mathrm{1}}{{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}}\::\:\frac{\mathrm{1}+{x}\left(\sqrt{{x}}−{x}^{\mathrm{2}} \right)}{\:\sqrt{{x}}\:−{x}^{\mathrm{2}} }\:\:\: \\ $$$$\frac{\sqrt{{x}}\:+\mathrm{1}}{{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}}×\frac{\sqrt{{x}}\:−{x}^{\mathrm{2}} }{\mathrm{1}+{x}\sqrt{{x}}−{x}^{\mathrm{3}} } \\ $$$$\frac{\left(\sqrt{{x}}\:+\mathrm{1}\right)\left(\sqrt{{x}}\:−{x}^{\mathrm{2}} \right)}{\left({x}\sqrt{{x}}\:+{x}+\sqrt{{x}}\right)\left(\mathrm{1}+{x}\sqrt{{x}}−{x}^{\mathrm{3}} \right)} \\ $$$$=\frac{\overset{×} {\left({x}−{x}^{\mathrm{2}} +\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}}\right)}}{\overset{×} {\left({x}−{x}^{\mathrm{2}} +\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}}\right)}\left(\mathrm{1}−{x}^{\mathrm{3}} +{x}\sqrt{{x}}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{3}} +{x}\sqrt{{x}}\right)}×\frac{\left(\mathrm{1}−{x}^{\mathrm{3}} −{x}\sqrt{{x}}\right)}{\left(\mathrm{1}−{x}^{\mathrm{3}} −{x}\sqrt{{x}}\right)} \\ $$$$=\frac{\left(\mathrm{1}−{x}^{\mathrm{3}} −{x}\sqrt{{x}}\right)}{\left.\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\mathrm{2}} −{x}^{\mathrm{3}} \right)}=\frac{\left(\mathrm{1}−{x}^{\mathrm{3}} −{x}\sqrt{{x}}\right)}{\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{6}} −{x}^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{1}−{x}^{\mathrm{3}} −{x}\sqrt{{x}}\right)}{{x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{3}} +\mathrm{1}} \\ $$
Commented by bemath last updated on 19/Aug/20
Commented by bemath last updated on 19/Aug/20
$${but}\:{nothing}\:{answer}\: \\ $$
Commented by Rasheed.Sindhi last updated on 19/Aug/20
$$\mathrm{Ok}\:\mathrm{sir},\:{afterall}\:{it}'{s}\:{a}\:{matter}\:{of} \\ $$$${convention}\:{only}! \\ $$
Answered by bemath last updated on 19/Aug/20
Answered by 1549442205PVT last updated on 19/Aug/20
![Put (((√x) +1)/(x(√x) +x+(√x))) : (1/( (√x) −x^2 )) + x =P Since x(√x) +x+(√x)=(√x)(x+(√x)+1) (√x)−x^2 =(√x)(1−((√x))^3 )=(√x)(1−(√x))(1+(√x)+x), P=[(((√x)+1)/( (√x)(1+(√x)+x)))×(√x)(1−(√x))(1+(√x)+x)]+x =((√x)+1)(1−(√x))+x=1−x+x=1 Thus,P=1](https://www.tinkutara.com/question/Q108756.png)
$$\mathrm{Put}\:\:\frac{\sqrt{{x}}\:+\mathrm{1}}{{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}}\::\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:−{x}^{\mathrm{2}} }\:+\:{x}\:=\mathrm{P} \\ $$$$\mathrm{Since}\:{x}\sqrt{{x}}\:+{x}+\sqrt{{x}}=\sqrt{\mathrm{x}}\left(\mathrm{x}+\sqrt{\mathrm{x}}+\mathrm{1}\right) \\ $$$$\sqrt{\mathrm{x}}−\mathrm{x}^{\mathrm{2}} =\sqrt{\mathrm{x}}\left(\mathrm{1}−\left(\sqrt{\mathrm{x}}\right)^{\mathrm{3}} \right)=\sqrt{\mathrm{x}}\left(\mathrm{1}−\sqrt{\mathrm{x}}\right)\left(\mathrm{1}+\sqrt{\mathrm{x}}+\mathrm{x}\right), \\ $$$$\mathrm{P}=\left[\frac{\sqrt{\mathrm{x}}+\mathrm{1}}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\sqrt{\mathrm{x}}+\mathrm{x}\right)}×\sqrt{\mathrm{x}}\left(\mathrm{1}−\sqrt{\mathrm{x}}\right)\left(\mathrm{1}+\sqrt{\mathrm{x}}+\mathrm{x}\right)\right]+\mathrm{x} \\ $$$$=\left(\sqrt{\mathrm{x}}+\mathrm{1}\right)\left(\mathrm{1}−\sqrt{\mathrm{x}}\right)+\mathrm{x}=\mathrm{1}−\mathrm{x}+\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{Thus},\mathrm{P}=\mathrm{1} \\ $$
Commented by bemath last updated on 19/Aug/20
$${jooss} \\ $$