Question Number 173504 by AgniMath last updated on 12/Jul/22 Answered by behi834171 last updated on 12/Jul/22 $${A}=\sqrt{{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}\Rightarrow{A}^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}}+\mathrm{2}=\mathrm{4}\Rightarrow{A}=\mathrm{2}\:\:.\:\:\blacksquare \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 173500 by Shrinava last updated on 12/Jul/22 $$\mathrm{Find}\:\mathrm{without}\:\mathrm{any}\:\mathrm{software}: \\ $$$$\Omega\:=\:\int\:\left(\mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\right)\left(\mathrm{1}\:−\:\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{sin}\left(\mathrm{ln}\left(\mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\mathrm{dx} \\ $$ Answered by thfchristopher last updated on 12/Jul/22 $$\mathrm{Let}\:{u}={x}+\frac{\mathrm{5}}{{x}} \\ $$$${du}=\left(\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}}…
Question Number 173503 by Shrinava last updated on 12/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173497 by SIENSE last updated on 12/Jul/22 $$\:\:\: \\ $$$$\:\:{Bonus}\:{du}\:{Mardi}\:\mathrm{12}/\mathrm{07}/\mathrm{2022} \\ $$$$\:\:\:\:{I}=\:\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)}{dx}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:{J}\:\:=\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$\:{I}\:=\:\int_{\mathrm{0}}…
Question Number 173489 by AgniMath last updated on 12/Jul/22 Answered by Rasheed.Sindhi last updated on 12/Jul/22 $${x}+{y}+{z}=\mathrm{0}\:;\:\frac{{x}^{\mathrm{2}} }{{yz}}+\frac{{y}^{\mathrm{2}} }{{zx}}+\frac{{z}^{\mathrm{2}} }{{xy}}=? \\ $$$$\:\frac{{x}^{\mathrm{2}} }{{yz}}+\frac{{y}^{\mathrm{2}} }{{zx}}+\frac{{z}^{\mathrm{2}} }{{xy}}=\:\frac{{x}^{\mathrm{3}}…
Question Number 173485 by mnjuly1970 last updated on 12/Jul/22 $$ \\ $$$$\:\:\:\:\:{Solve}\:\:\:\:\:\:\:\left({x}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\:\:\:\:{x}\mathrm{2}^{\:\frac{\mathrm{1}}{{x}}} \:+\frac{\mathrm{1}}{{x}}\:\mathrm{2}^{\:{x}} =\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:−{Source}:\:{L}.{Panaitopol} \\ $$ Answered by aleks041103 last updated…
Question Number 107947 by I want to learn more last updated on 13/Aug/20 Answered by mr W last updated on 13/Aug/20 $$\left.{a}\right) \\ $$$$\mathrm{10}!=\mathrm{3628800} \\…
Question Number 42408 by LYCON TRIX last updated on 25/Aug/18 $$\sqrt{{a}−\sqrt{{a}+{x}\:}}+\:\sqrt{{a}+\sqrt{{a}−{x}}\:}\:=\:\mathrm{2}{x} \\ $$$${Solve}\:{for}\:{x}\:{in}\:{terms}\:{of}\:{a} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18 $${really}\:{excellent}\:{problem}…{searching}\:{way}\:{to}\:{solve} \\ $$…
Question Number 107945 by bemath last updated on 13/Aug/20 $$\:\:\:\:\:\:\frac{\circ\mathbb{B}{e}\mathbb{M}{ath}\circ}{\wedge\smile\wedge} \\ $$$$\:\:\:\begin{cases}{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\mathrm{23}}\\{{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:?}\end{cases} \\ $$ Answered by $@y@m last updated on 13/Aug/20…
Question Number 173445 by AgniMath last updated on 11/Jul/22 Answered by Rasheed.Sindhi last updated on 11/Jul/22 $$\frac{{a}}{\mathrm{3}}=\frac{{a}+{b}}{\mathrm{4}}=\frac{{a}+{b}+{c}}{\mathrm{5}}=\frac{{a}+{b}+{c}+{d}}{\mathrm{6}}={k} \\ $$$${a}=\mathrm{3}{k} \\ $$$${a}+{b}=\mathrm{4}{k}\Rightarrow{b}=\mathrm{4}{k}−\mathrm{3}{k}={k} \\ $$$${a}+{b}+{c}=\mathrm{5}{k}\Rightarrow\mathrm{4}{k}+{c}=\mathrm{5}{k}\Rightarrow{c}={k} \\ $$$${a}+{b}+{c}+{d}=\mathrm{6}{k}\Rightarrow\mathrm{5}{k}+{d}=\mathrm{6}{k}\Rightarrow{d}={k}…