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Question Number 200886 by essaad last updated on 26/Nov/23
resoudre dans R : ((3+x))^(1/3) −((3−x))^(1/3) =((9−x^2 ))^(1/3)
$${resoudre}\:{dans}\:{R}\:: \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:=\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$
Answered by Frix last updated on 26/Nov/23
a^(1/3) −b^(1/3) =c^(1/3) a−b−3a^(1/3) b^(1/3) (a^(1/3) −b^(1/3) )=c a−b−3a^(1/3) b^(1/3) c^(1/3) =c 27abc=(a−b−c)^3 c=ab 27a^2 b^2 =(a−b−ab)^3 a=3+x∧b=3−x 27(x−3)^2 (x+3)^2 =(x^2 +2x−9)^3 x^6 +6x^5 −42x^4 −100x^3 +621x^2 +486x−2916=0 No exact solution possible x≈−8.70670378^★ ∨x≈2.73590400 ^★ only if we use ((−r))^(1/3) =−(r)^(1/3)
$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}−{b}−\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)={c} \\ $$$${a}−{b}−\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} {c}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c} \\ $$$$\mathrm{27}{abc}=\left({a}−{b}−{c}\right)^{\mathrm{3}} \\ $$$${c}={ab} \\ $$$$\mathrm{27}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\left({a}−{b}−{ab}\right)^{\mathrm{3}} \\ $$$${a}=\mathrm{3}+{x}\wedge{b}=\mathrm{3}−{x} \\ $$$$\mathrm{27}\left({x}−\mathrm{3}\right)^{\mathrm{2}} \left({x}+\mathrm{3}\right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{9}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{6}} +\mathrm{6}{x}^{\mathrm{5}} −\mathrm{42}{x}^{\mathrm{4}} −\mathrm{100}{x}^{\mathrm{3}} +\mathrm{621}{x}^{\mathrm{2}} +\mathrm{486}{x}−\mathrm{2916}=\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible} \\ $$$${x}\approx−\mathrm{8}.\mathrm{70670378}^{\bigstar} \vee{x}\approx\mathrm{2}.\mathrm{73590400} \\ $$$$\:^{\bigstar} \:\mathrm{only}\:\mathrm{if}\:\mathrm{we}\:\mathrm{use}\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$
Commented by essaad last updated on 26/Nov/23
thanks sir
$${thanks}\:{sir}\: \\ $$
Answered by Rasheed.Sindhi last updated on 26/Nov/23
Another way ((3+x))^(1/3) −((3−x))^(1/3) −((9−x^2 ))^(1/3) =0 a=((3+x))^(1/3) ,b= ((3−x))^(1/3) , c=((9−x^2 ))^(1/3) a−b−c=0 determinant (((a^3 −b^3 −c^3 −3abc=(a−b−c)(a^2 +b^2 +c^2 +ab−bc+ca)))) a^3 −b^3 −c^3 −3abc=0 (3+x)−(3−x)−(9−x^2 )−3(((3+x))^(1/3) )(((3−x))^(1/3) )(((9−x^2 ))^(1/3) )=0 x^2 +2x−9=3(((9−x^2 )^2 ))^(1/3) =0 (x^2 +2x−9)^3 =27(x^2 −9)^2 x^6 +6x^5 −42x^4 −100x^3 +621x^2 +486x−2916=0 x≈−8.70670377676 , 2.73590400403
$$\mathbb{A}\mathrm{nother}\:\mathrm{way} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${a}=\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:,{b}=\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:\:,\:{c}=\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${a}−{b}−{c}=\mathrm{0} \\ $$$$\begin{array}{|c|}{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} −\mathrm{3}{abc}=\left({a}−{b}−{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}−{bc}+{ca}\right)}\\\hline\end{array} \\ $$$$\:\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{0} \\ $$$$\left(\mathrm{3}+{x}\right)−\left(\mathrm{3}−{x}\right)−\left(\mathrm{9}−{x}^{\mathrm{2}} \right)−\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} }\:\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{9}=\mathrm{3}\sqrt[{\mathrm{3}}]{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{9}\right)^{\mathrm{3}} =\mathrm{27}\left({x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{6}} +\mathrm{6}{x}^{\mathrm{5}} −\mathrm{42}{x}^{\mathrm{4}} −\mathrm{100}{x}^{\mathrm{3}} +\mathrm{621}{x}^{\mathrm{2}} +\mathrm{486}{x}−\mathrm{2916}=\mathrm{0} \\ $$$${x}\approx−\mathrm{8}.\mathrm{70670377676}\:,\:\mathrm{2}.\mathrm{73590400403} \\ $$
Answered by mr W last updated on 26/Nov/23
((3+x))^(1/3) −((3−x))^(1/3) =(((3+x)(3−x)))^(1/3) say u=((3+x))^(1/3) ,v=((3−x))^(1/3) u−v=uv ⇒v=(u/(1+u)) u^3 +v^3 =6 u^3 +(u^3 /((1+u)^3 ))=6 u^6 +3u^5 +3u^4 −4u^3 −18u^2 −18u−6=0 ⇒u≈−1.787016, 1.790059 ⇒x=u^3 −3≈−8.706704, ≈2.735906
$$\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}−\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}=\sqrt[{\mathrm{3}}]{\left(\mathrm{3}+{x}\right)\left(\mathrm{3}−{x}\right)} \\ $$$${say}\:{u}=\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}},{v}=\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}} \\ $$$${u}−{v}={uv}\:\Rightarrow{v}=\frac{{u}}{\mathrm{1}+{u}} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\mathrm{6} \\ $$$${u}^{\mathrm{3}} +\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} }=\mathrm{6} \\ $$$${u}^{\mathrm{6}} +\mathrm{3}{u}^{\mathrm{5}} +\mathrm{3}{u}^{\mathrm{4}} −\mathrm{4}{u}^{\mathrm{3}} −\mathrm{18}{u}^{\mathrm{2}} −\mathrm{18}{u}−\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow{u}\approx−\mathrm{1}.\mathrm{787016},\:\mathrm{1}.\mathrm{790059} \\ $$$$\Rightarrow{x}={u}^{\mathrm{3}} −\mathrm{3}\approx−\mathrm{8}.\mathrm{706704},\:\approx\mathrm{2}.\mathrm{735906} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Nov/23
((3+x))^(1/3) −((3−x))^(1/3) =((9−x^2 ))^(1/3) Dividing by ((9−x^2 ))^(1/3) : (1/( ((3−x))^(1/3) ))−(1/( ((3+x))^(1/3) ))=1 a=(1/( ((3−x))^(1/3) )) , b=(1/( ((3+x))^(1/3) )) ⇒ a−b−1=0 determinant (((a^3 −b^3 −1−3ab=(a−b−1)_(⇒a^3 −b^3 −1−3ab=0 ) (a^2 +b^2 +1+ab−b+a)))) (1/( 3−x ))−(1/(3+x))−1−3((1/( ((3−x))^(1/3) )))((1/( ((3+x))^(1/3) )))=0 (1/( 3−x ))−(1/(3+x))−1=3((1/( ((3−x))^(1/3) )))((1/( ((3+x))^(1/3) ))) ((3+x−3+x−9+x^2 )/(9−x^2 ))=(3/( ((9−x^2 ))^(1/3) )) x^2 +2x−9=3(9−x^2 )^(2/3) (x^2 +2x−9)^3 =27(9−x^2 )^2 x^6 +6x^5 −42x^4 −100x^3 +621x^2 +486x−2916=0 x≈−8.70670377676 , 2.73590400403
$$\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:=\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$$\mathcal{D}{ividing}\:{by}\:\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} }\:: \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:}=\mathrm{1} \\ $$$${a}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:}\:,\:{b}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:}\:\Rightarrow \\ $$$${a}−{b}−\mathrm{1}=\mathrm{0} \\ $$$$\begin{array}{|c|}{\underset{\Rightarrow{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −\mathrm{1}−\mathrm{3}{ab}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −\mathrm{1}−\mathrm{3}{ab}=\left({a}−{b}−\mathrm{1}\right)}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{1}+{ab}−{b}+{a}\right)}\\\hline\end{array} \\ $$$$\frac{\mathrm{1}}{\:\mathrm{3}−{x}\:}−\frac{\mathrm{1}}{\mathrm{3}+{x}}−\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\:\mathrm{3}−{x}\:}−\frac{\mathrm{1}}{\mathrm{3}+{x}}−\mathrm{1}=\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:}\right) \\ $$$$\frac{\mathrm{3}+{x}−\mathrm{3}+{x}−\mathrm{9}+{x}^{\mathrm{2}} }{\mathrm{9}−{x}^{\mathrm{2}} }=\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{9}=\mathrm{3}\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{9}\right)^{\mathrm{3}} =\mathrm{27}\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{6}} +\mathrm{6}{x}^{\mathrm{5}} −\mathrm{42}{x}^{\mathrm{4}} −\mathrm{100}{x}^{\mathrm{3}} +\mathrm{621}{x}^{\mathrm{2}} +\mathrm{486}{x}−\mathrm{2916}=\mathrm{0} \\ $$$${x}\approx−\mathrm{8}.\mathrm{70670377676}\:,\:\mathrm{2}.\mathrm{73590400403} \\ $$

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