Question Number 90139 by jagoll last updated on 21/Apr/20 $$\mathrm{x}\:=\:\mathrm{2021}^{\mathrm{3}} −\mathrm{2019}^{\mathrm{3}} \\ $$$$\sqrt{\frac{\mathrm{x}−\mathrm{2}}{\mathrm{6}}}\:=\:? \\ $$ Commented by jagoll last updated on 21/Apr/20 $$\mathrm{x}\:=\:\left(\mathrm{p}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{p}^{\mathrm{3}} \:,\:\mathrm{p}\:=\:\mathrm{2019}…
Question Number 155667 by amin96 last updated on 03/Oct/21 Commented by Tawa11 last updated on 03/Oct/21 $$\mathrm{weldone}\:\mathrm{sir} \\ $$ Commented by amin96 last updated on…
Question Number 155650 by mathdanisur last updated on 03/Oct/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}}\:+\:\frac{\mathrm{a}}{\mathrm{x}}\:=\:\sqrt{\frac{\mathrm{x}-\mathrm{b}}{\mathrm{x}-\mathrm{a}}}\:+\:\frac{\mathrm{b}}{\mathrm{x}} \\ $$$$\mathrm{a};\mathrm{b}\in\mathbb{R}\:\:\mathrm{and}\:\:\mathrm{a}\neq\mathrm{b} \\ $$ Answered by som(math1967) last updated on 03/Oct/21 $$\sqrt{\frac{{x}−{a}}{{x}−{b}}}−\sqrt{\frac{{x}−{b}}{{x}−{a}}}=\frac{{b}−{a}}{{x}} \\…
Question Number 155653 by mathdanisur last updated on 03/Oct/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{1}\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{18}\:\Sigma\:\mathrm{ab}\:+\:\mathrm{45}\:\Sigma\:\mathrm{a}^{\mathrm{2}} \mathrm{b}\:\leqslant\:\mathrm{11} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 24576 by math solver last updated on 21/Nov/17 Commented by math solver last updated on 21/Nov/17 $$\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:? \\ $$ Answered by mrW1 last…
Question Number 90103 by A8;15: last updated on 21/Apr/20 Commented by john santu last updated on 21/Apr/20 $${translate} \\ $$ Terms of Service Privacy Policy…
Question Number 24565 by ajfour last updated on 21/Nov/17 $$\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{3}} +\boldsymbol{{bx}}^{\mathrm{2}} +\boldsymbol{{cx}}+\boldsymbol{{d}}\:,\:{then} \\ $$$${prove}\:{that}\:{the}\:{equation}\:{y}=\mathrm{0} \\ $$$${has}\:{only}\:{one}\:{real}\:{root}\:{if} \\ $$$$\:\boldsymbol{{a}}\left[\left(\mathrm{9}\boldsymbol{{ad}}−\boldsymbol{{bc}}\right)^{\mathrm{2}} −\mathrm{4}\left(\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{ac}}\right)\left(\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{bd}}\right)\right] \\ $$$$\:\:\:\:>\:\mathrm{0}\:\:\:\:\:{provided}\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \:>\:\mathrm{3}\boldsymbol{{ac}}\:. \\…
Question Number 90097 by jagoll last updated on 21/Apr/20 $$\left(\sqrt{\mathrm{3}+\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:+\left(\sqrt{\mathrm{3}−\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:=\:\mathrm{6} \\ $$ Commented by john santu last updated on 21/Apr/20 $$\left(\mathrm{3}+\sqrt{\mathrm{8}}\right)\left(\mathrm{3}−\sqrt{\mathrm{8}}\right)=\mathrm{1} \\ $$$$\mathrm{3}−\sqrt{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{8}}}\:…
Question Number 90092 by jagoll last updated on 21/Apr/20 $$\mathrm{G}\left(\sqrt{\mathrm{x}+\mathrm{5}}\right)\:=\:\mathrm{x} \\ $$$$\mathrm{G}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{x}^{\mathrm{a}} −\mathrm{b} \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b}\: \\ $$ Commented by john santu last updated on…
Question Number 155621 by mathdanisur last updated on 02/Oct/21 Answered by ghimisi last updated on 03/Oct/21 $$\frac{{a}+{b}+{c}}{\mathrm{6}}=\mathrm{1}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{a}}\right)^{\frac{{a}}{\mathrm{6}}} \left(\frac{{b}+{e}}{{b}}\right)^{\frac{{b}}{\mathrm{6}}} \left(\frac{{c}+{f}}{{c}}\right)^{\frac{{c}}{\mathrm{6}}} \leqslant\frac{{a}}{\mathrm{6}}\centerdot\frac{{a}+{d}}{{a}}+\frac{{b}}{\mathrm{6}}\centerdot\frac{{b}+{e}}{{b}}+\frac{{c}}{\mathrm{6}}\centerdot\frac{{c}+{f}}{{c}}=\mathrm{2}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{a}}\right)^{{a}} \left(\frac{{b}+{c}}{{b}}\right)^{{b}}…