Question Number 213735 by Spillover last updated on 15/Nov/24
Answered by issac last updated on 15/Nov/24
$$\mathrm{derivate}\:\mathrm{each}\:\mathrm{side} \\ $$$$\mathrm{0}=\mathrm{0}\:\:\rightarrow\:\mathrm{true} \\ $$$$\mathrm{Q}.\mathrm{E}.\mathrm{D} \\ $$
Commented by Frix last updated on 15/Nov/24
![Great, with this method everything is true! • 5=−7 true because (d/dx)[5]=0∧(d/dx)[−7]=0 • a^2 +b^2 =c^2 true because (d/dx)[a^2 +b^2 ]=0∧(d/dx)[c^2 ]=0 • ((πln 2)/γ)=∫_0 ^∞ t^(sin t) ln (1−t) dt true because (d/dx)[((πln 2)/γ)]=0∧(d/dx)[∫_0 ^∞ t^(sin t) ln (1−t) dt]=0](https://www.tinkutara.com/question/Q213739.png)
$$\mathrm{Great},\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{everything}\:\mathrm{is}\:\mathrm{true}! \\ $$$$\bullet\:\mathrm{5}=−\mathrm{7}\:\mathrm{true}\:\mathrm{because} \\ $$$$\:\:\:\:\:\frac{{d}}{{dx}}\left[\mathrm{5}\right]=\mathrm{0}\wedge\frac{{d}}{{dx}}\left[−\mathrm{7}\right]=\mathrm{0} \\ $$$$\bullet\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\mathrm{true}\:\mathrm{because} \\ $$$$\:\:\:\:\:\frac{{d}}{{dx}}\left[{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right]=\mathrm{0}\wedge\frac{{d}}{{dx}}\left[{c}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\bullet\:\frac{\pi\mathrm{ln}\:\mathrm{2}}{\gamma}=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{\mathrm{sin}\:{t}} \mathrm{ln}\:\left(\mathrm{1}−{t}\right)\:{dt}\:\mathrm{true}\:\mathrm{because} \\ $$$$\:\:\:\:\:\frac{{d}}{{dx}}\left[\frac{\pi\mathrm{ln}\:\mathrm{2}}{\gamma}\right]=\mathrm{0}\wedge\frac{{d}}{{dx}}\left[\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{\mathrm{sin}\:{t}} \mathrm{ln}\:\left(\mathrm{1}−{t}\right)\:{dt}\right]=\mathrm{0} \\ $$
Commented by Ghisom last updated on 15/Nov/24
$$\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{useful}\:\mathrm{for}\:\mathrm{solving}\:\mathrm{this}: \\ $$$${x}^{\mathrm{5}} +{ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\mathrm{simply}\:\mathrm{differentiate}\:\mathrm{3}\:\mathrm{times}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{60}{x}^{\mathrm{2}} +\mathrm{24}{ax}+\mathrm{6}{b}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{you}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{general}\:\mathrm{quintic}\:\mathrm{using} \\ $$$$\mathrm{only}\:\mathrm{square}\:\mathrm{roots} \\ $$$$\mathrm{the}\:\mathrm{Nobel}\:\mathrm{Prize}\:\mathrm{awaits}\:\mathrm{you}! \\ $$
Answered by Spillover last updated on 15/Nov/24
Answered by Spillover last updated on 15/Nov/24
