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Question Number 184091 by Ib last updated on 02/Jan/23

Commented by Ib last updated on 02/Jan/23

$$jaibesoindaide$$$$$$$$$$

Answered by CElcedricjunior last updated on 02/Jan/23

$$A={\left(2\mathit{x}\right)}^{\frac{3}{2}}=\sqrt[2]{{\left(2\mathit{x}\right)}^3}$$$${\mathit{A}}^2=8{\mathit{x}}^3$$$$\mathit{B}={\left(2x^3\right)}^{\frac{1}{2}}+{\left(2x^3\right)}^{\frac{1}{2}}=2{\left(2{\mathit{x}}^3\right)}^{\frac{1}{2}}=\sqrt{8{\mathit{x}}^3}$$$$=>{\mathit{B}}^2=8{\mathit{x}}^3$$$$<=>{\mathit{A}}^2-{\mathit{B}}^2=8x^3-8x^3=0$$$$=>{\mathit{A}}^2={\mathit{B}}^2=>A=\mathit{B}$$$$2){(1+\frac{\mathit{t}}{100})}^{10}=3=>\frac{\mathit{t}}{100}<<<<1$$$$=>{(1+\frac{\mathit{t}}{100})}^{10}=3\mathit{o}\mathit{r}{(1+\mathit{a})}^{\mathit{n}}=1+\mathit{n}\mathit{a}\mathit{p}\mathit{o}\mathit{u}\mathit{r}\mathit{a}<<1$$$$=>1+\frac{10\mathit{t}}{100}=3$$$$=>\frac{\mathit{t}}{10}=2=>\mathit{t}=20$$$$==========================$$$$......................lecelebrecedricjunior........$$$$========================$$$$\mathit{y}\left(\mathit{x}\right)={\mathit{a}\mathit{e}}^{-6\mathit{x}}+{\mathit{b}\mathit{e}}^{-2\mathit{x}}+\mathit{c}$$$$\mathit{a};\mathit{b}\mathit{e}\mathit{t}\mathit{c}\in \mathbb{R}$$$$$$

Answered by manxsol last updated on 02/Jan/23

$$B={\left(2x^3\right)}^{\frac{1}{2}}+{\left(2x^3\right)}^{\frac{1}{2}}$$$$B=2^{\frac{1}{2}}{x}^{\frac{3}{2}}+2^{\frac{1}{2}}{x}^{\frac{3}{2}}=m+m=2m$$$$B=2.2^{\frac{1}{2}}{x}^{\frac{3}{2}}$$$$B=2^{\frac{1}{1}+\frac{1}{2}}{x}^{\frac{3}{2}}$$$$B=2^{\frac{3}{2}}.x^{\frac{3}{2}}$$$$B={\left(2x\right)}^{\frac{3}{2}}$$$$ensuite$$$$A=B={\left(2x\right)}^{\frac{3}{2}}$$$$$$

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