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Question-184091

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Question Number 184091 by Ib last updated on 02/Jan/23
Commented by Ib last updated on 02/Jan/23
j ai besoin d aide
$${j}\:{ai}\:{besoin}\:{d}\:{aide} \\ $$$$ \\ $$$$ \\ $$
Answered by CElcedricjunior last updated on 02/Jan/23
A=(2x)^(3/2) =(((2x)^3 ))^(1/2) A^2 =8x^3 B=(2x^3 )^(1/2) +(2x^3 )^(1/2) =2(2x^3 )^(1/2) =(√(8x^3 )) =>B^2 =8x^3 <=>A^2 −B^2 =8x^3 −8x^3 =0 =>A^2 =B^2 =>A=B 2)(1+(t/(100)))^(10) =3=>(t/(100))<<<<1 =>(1+(t/(100)))^(10) =3 or (1+a)^n =1+na pour a<<1 =>1+((10t)/(100))=3 =>(t/(10))=2=>t=20 ========================== ......................le celebre cedric junior........ ======================== y(x)=ae^(−6x) +be^(−2x) +c a;bet c∈R
$${A}=\left(\mathrm{2}\boldsymbol{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\sqrt[{\mathrm{2}}]{\left(\mathrm{2}\boldsymbol{{x}}\right)^{\mathrm{3}} } \\ $$$$\boldsymbol{{A}}^{\mathrm{2}} =\mathrm{8}\boldsymbol{{x}}^{\mathrm{3}} \\ $$$$\boldsymbol{{B}}=\left(\mathrm{2}{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left(\mathrm{2}{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{2}\left(\mathrm{2}\boldsymbol{{x}}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\mathrm{8}\boldsymbol{{x}}^{\mathrm{3}} } \\ $$$$=>\boldsymbol{{B}}^{\mathrm{2}} =\mathrm{8}\boldsymbol{{x}}^{\mathrm{3}} \\ $$$$<=>\boldsymbol{{A}}^{\mathrm{2}} −\boldsymbol{{B}}^{\mathrm{2}} =\mathrm{8}{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{3}} =\mathrm{0} \\ $$$$=>\boldsymbol{{A}}^{\mathrm{2}} =\boldsymbol{{B}}^{\mathrm{2}} =>{A}=\boldsymbol{{B}} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{1}+\frac{\boldsymbol{{t}}}{\mathrm{100}}\right)^{\mathrm{10}} =\mathrm{3}=>\frac{\boldsymbol{{t}}}{\mathrm{100}}<<<<\mathrm{1} \\ $$$$=>\left(\mathrm{1}+\frac{\boldsymbol{{t}}}{\mathrm{100}}\right)^{\mathrm{10}} =\mathrm{3}\:\:\boldsymbol{{or}}\:\left(\mathrm{1}+\boldsymbol{{a}}\right)^{\boldsymbol{{n}}} =\mathrm{1}+\boldsymbol{{na}}\:\boldsymbol{{pour}}\:\boldsymbol{{a}}<<\mathrm{1} \\ $$$$=>\mathrm{1}+\frac{\mathrm{10}\boldsymbol{{t}}}{\mathrm{100}}=\mathrm{3} \\ $$$$=>\frac{\boldsymbol{{t}}}{\mathrm{10}}=\mathrm{2}=>\boldsymbol{{t}}=\mathrm{20} \\ $$$$========================== \\ $$$$………………….{le}\:{celebre}\:{cedric}\:{junior}…….. \\ $$$$======================== \\ $$$$\boldsymbol{{y}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ae}}^{−\mathrm{6}\boldsymbol{{x}}} +\boldsymbol{{be}}^{−\mathrm{2}\boldsymbol{{x}}} +\boldsymbol{{c}}\:\: \\ $$$$\boldsymbol{{a}};\boldsymbol{{bet}}\:\boldsymbol{{c}}\in\mathbb{R} \\ $$$$ \\ $$
Answered by manxsol last updated on 02/Jan/23
B=(2x^3 )^(1/2) +(2x^3 )^(1/2) B=2^(1/2) x^(3/2) +2^(1/2) x^(3/2) =m+m=2m B=2.2^(1/2) x^(3/2) B=2^((1/1)+(1/2)) x^(3/2) B=2^(3/2) .x^(3/2) B=(2x)^(3/2) ensuite A=B=(2x)^(3/2)
$${B}=\left(\mathrm{2}{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left(\mathrm{2}{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${B}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}} ={m}+{m}=\mathrm{2}{m} \\ $$$${B}=\mathrm{2}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${B}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${B}=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} .{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${B}=\left(\mathrm{2}{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${ensuite} \\ $$$${A}={B}=\left(\mathrm{2}{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$

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