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Question Number 140654 by muallim_riyoziyot last updated on 11/May/21
Answered by MJS_new last updated on 11/May/21
for x∈R ((−x))^(1/3) =−(x)^(1/3) (2((2x+1))^(1/3) =x^3 −1)^3 8(2x+1)=(x^3 −1) x^9 −3x^6 +3x^3 −16x−9=0 2((2x+1))^(1/3) =x^3 −1 x=((t^3 −1)/2) 2t=(((t^3 −1)/2))^3 −1 t^9 −3t^6 +3t^3 −16t−9=0 ⇒ x=t ⇒ x=((x^3 −1)/2) x^3 −2x−1=0 (x+1)(x^2 −x−1)=0 x_1 =−1 x_2 =((1−(√5))/2) x_3 =((1+(√5))/2)
$$\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\sqrt[{\mathrm{3}}]{−{x}}=−\sqrt[{\mathrm{3}}]{{x}} \\ $$$$ \\ $$$$\left(\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}={x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\mathrm{8}\left(\mathrm{2}{x}+\mathrm{1}\right)=\left({x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$${x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{3}} −\mathrm{16}{x}−\mathrm{9}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}={x}^{\mathrm{3}} −\mathrm{1} \\ $$$${x}=\frac{{t}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{t}=\left(\frac{{t}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{1} \\ $$$${t}^{\mathrm{9}} −\mathrm{3}{t}^{\mathrm{6}} +\mathrm{3}{t}^{\mathrm{3}} −\mathrm{16}{t}−\mathrm{9}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:{x}={t}\:\Rightarrow\:{x}=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by phanphuoc last updated on 11/May/21
y=((2x+1))^(1/3) →y^3 =2x+1(1),and x^3 =2y+1(2) (1)−(2)→y^3 −x^3 =2(y−x)→(y−x)(y^2 +xy+y^2 +2)=0 ∴ if x=y→x^3 =2x+1→ez. if x^2 +xy+y^2 +2=(x+y/2)^2 +3y^2 /4+2>=2
$${y}=\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}\rightarrow{y}^{\mathrm{3}} =\mathrm{2}{x}+\mathrm{1}\left(\mathrm{1}\right),{and}\:{x}^{\mathrm{3}} =\mathrm{2}{y}+\mathrm{1}\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\rightarrow{y}^{\mathrm{3}} −{x}^{\mathrm{3}} =\mathrm{2}\left({y}−{x}\right)\rightarrow\left({y}−{x}\right)\left({y}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$$\therefore\:{if}\:{x}={y}\rightarrow{x}^{\mathrm{3}} =\mathrm{2}{x}+\mathrm{1}\rightarrow{ez}. \\ $$$${if}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} +\mathrm{2}=\left({x}+{y}/\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} /\mathrm{4}+\mathrm{2}>=\mathrm{2} \\ $$$$ \\ $$
Answered by john_santu last updated on 11/May/21
((x^3 −1)/2) = ((2x+1))^(1/(3 )) = y ⇒ x^3 = 2y +1 , y = ((2x+1))^(1/(3 )) ⇒y^3 = 2x+1 ⇒x^3 −y^3 = 2(y−x) ⇒x^3 −y^3 +2x−2y=0 ⇒(x−y)(x^2 +xy+y^2 )+2(x−y)=0 ⇒(x−y)(x^2 +xy+y^2 +2)=0 ⇒x = y ; ((x^3 −1)/2) = x ⇒x^3 −2x−1 = 0 ⇒(x+1)(x^2 −x−1)=0 ⇒(x+1)(x^2 −x−1) =0 ⇒x=−1 ∧ x = ((1 ± (√5))/2)
$$\:\frac{{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}{x}+\mathrm{1}}\:=\:{y} \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:=\:\mathrm{2}{y}\:+\mathrm{1}\:,\:{y}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\Rightarrow{y}^{\mathrm{3}} \:=\:\mathrm{2}{x}+\mathrm{1}\: \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{y}^{\mathrm{3}} \:=\:\mathrm{2}\left({y}−{x}\right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{y}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{2}{y}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)+\mathrm{2}\left({x}−{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}\:=\:{y}\:;\:\frac{{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}}\:=\:{x}\: \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)=\mathrm{0}\: \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\:=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{1}\:\wedge\:{x}\:=\:\frac{\mathrm{1}\:\pm\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 11/May/21
((2x+1))^(1/3) =((x^3 −1)/2) let f(x)=((2x+1))^(1/3) ⇒f^(−1) (x)=((x^3 −1)/2) ⇒f(x)=f^(−1) (x) ⇒f(x)=f^(−1) (x)=x ⇒((x^3 −1)/2)=x ⇒x^3 −2x−1=0 ⇒(x+1)(x^2 −x−1)=0 ⇒x=−1, ((1±(√5))/2)
$$\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:{f}\left({x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)={x} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{1},\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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