Question Number 65664 by mathmax by abdo last updated on 01/Aug/19
![solve (((√(1−x))−(√(2x+1)))/( (√(1−x))+(√(2x+1)))) =((x+1)/3)](https://www.tinkutara.com/question/Q65664.png)
$${solve}\:\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}\:=\frac{{x}+\mathrm{1}}{\mathrm{3}} \\ $$
Answered by MJS last updated on 01/Aug/19
![(((√(1−x))−(√(2x+1)))/( (√(1−x))+(√(2x+1))))=((((√(1−x))−(√(2x+1)))^2 )/(((√(1−x))+(√(2x+1)))((√(1−x))−(√(2x+1)))))= =((x+2−2(√(1−x))(√(2x+1)))/(−3x))=−((x+2)/(3x))+((2(√(1−x))(√(2x+1)))/(3x)) −((x+2)/(3x))+((2(√(1−x))(√(2x+1)))/(3x))=((x+1)/3) 2(√(1−x))(√(2x+1))=x^2 +2x+2 squaring x(x^3 +4x^2 +16x+4)=0 x_1 =0 but it′s false x^3 +4x^2 +16x+4=0 x=t−(4/3) t=((((170)/(27))+(2/9)(√(1713))))^(1/3) −((−((170)/(27))+(2/9)(√(1713))))^(1/3) x=−(4/3)+((((170)/(27))+(2/9)(√(1713))))^(1/3) −((−((170)/(27))+(2/9)(√(1713))))^(1/3) x≈−.266582](https://www.tinkutara.com/question/Q65671.png)
$$\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}=\frac{\left(\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)\left(\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)}= \\ $$$$=\frac{{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{−\mathrm{3}{x}}=−\frac{{x}+\mathrm{2}}{\mathrm{3}{x}}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}} \\ $$$$−\frac{{x}+\mathrm{2}}{\mathrm{3}{x}}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}}=\frac{{x}+\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2} \\ $$$$\mathrm{squaring} \\ $$$${x}\left({x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{false} \\ $$$${x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}={t}−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}} \\ $$$${x}=−\frac{\mathrm{4}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{170}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{1713}}} \\ $$$${x}\approx−.\mathrm{266582} \\ $$
Commented by mathmax by abdo last updated on 01/Aug/19
![thank you sir.](https://www.tinkutara.com/question/Q65672.png)
$${thank}\:{you}\:{sir}. \\ $$