Question Number 18864 by chernoaguero@gmail.com last updated on 31/Jul/17 Commented by mrW1 last updated on 31/Jul/17 $$\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(−\mathrm{2},\mathrm{1}\right),\left(\mathrm{1},−\mathrm{2}\right),\left(\mathrm{0},−\mathrm{3}\right),\left(−\mathrm{3},\mathrm{0}\right) \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on…
Question Number 84394 by jagoll last updated on 12/Mar/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\frac{\mathrm{2x}}{\mathrm{x}−\mathrm{2}}\:\leqslant\:\mid\mathrm{x}−\mathrm{3}\mid\: \\ $$ Commented by john santu last updated on 12/Mar/20 $$\left(\mathrm{i}\right)\:\mathrm{x}\:\geqslant\mathrm{3}\:\Rightarrow\:\frac{\mathrm{2x}}{\mathrm{x}−\mathrm{2}}\:\leqslant\:\mathrm{x}−\mathrm{3} \\ $$$$\Rightarrow\mathrm{2x}\:\leqslant\:\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{2}\right)…
Question Number 84393 by agniv123 last updated on 12/Mar/20 $${if}\:{x}^{{x}} .{y}^{{y}} .{z}^{{z}} ={x}^{{y}} .{y}^{{z}} .{z}^{{x}} ={x}^{{z}} .{y}^{{x}} .{z}^{{y}} \:{such}\:{that}\:{x},\:{y}\:{and}\:{z}\: \\ $$$${are}\:{positive}\:{intigers}\:{greater}\:{than}\:\mathrm{1} \\ $$$$,{what}\:{is}\:{the}\:{value}\:{of}\:{xyz}\:{and}\:{x}+{y}+{z}\:? \\ $$…
Question Number 84384 by M±th+et£s last updated on 12/Mar/20 $$\left[{x}\right]^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:\:,\:\forall{x}>\mathrm{0} \\ $$ Commented by mr W last updated on 12/Mar/20 $$\left[{x}\right]\leqslant{x} \\ $$$$\left[{x}\right]^{{x}} \leqslant{x}^{{x}}…
Question Number 149917 by mathdanisur last updated on 08/Aug/21 $$\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\boldsymbol{\pi}} {\int}}\frac{\mathrm{dt}}{\mathrm{4}\sqrt{\mathrm{2}}\:\mathrm{sin}\boldsymbol{\mathrm{t}}\:+\:\mathrm{6}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 08/Aug/21 $$\Psi=\int_{\mathrm{0}} ^{\mathrm{2}\pi}…
Question Number 84370 by Roland Mbunwe last updated on 12/Mar/20 $$\left.\mathrm{1}.\right)\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\:<\mathrm{5} \\ $$$$\left.\mathrm{2}.\right)\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\:+\:\mid\mathrm{2}−{x}\mid\:\leqslant\mathrm{8} \\ $$ Answered by john santu last updated on 12/Mar/20 $$\left.\mathrm{1}.\right)\:\left(\mathrm{i}\right)\:\mathrm{x}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{2x}\:<\:\mathrm{3}\:,\:\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}}\: \\…
Question Number 149889 by Samimsultani last updated on 08/Aug/21 Commented by bramlexs22 last updated on 08/Aug/21 $$\sqrt{\mathrm{a}+\mathrm{b}\sqrt{\mathrm{d}}}\:=\:\sqrt{\frac{\mathrm{a}+\mathrm{c}}{\mathrm{2}}}\:+\mathrm{sgn}\left(\mathrm{b}\right)\sqrt{\frac{\mathrm{a}−\mathrm{c}}{\mathrm{2}}} \\ $$$$\mathrm{where}\:\mathrm{c}=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{d}}\: \\ $$ Commented by…
Question Number 149876 by liberty last updated on 08/Aug/21 Answered by MJS_new last updated on 08/Aug/21 $${y}\geqslant\mathrm{0} \\ $$$${y}=\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\mathrm{squaring} \\ $$$${y}^{\mathrm{2}} =\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)}+\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right) \\…
Question Number 149883 by liberty last updated on 08/Aug/21 $$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rational}\:\mathrm{number} \\ $$ Answered by john_santu last updated on 08/Aug/21 $$\mathrm{L}{et}\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\: \\ $$$${We}\:{then}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{0} \\…
Question Number 149868 by mathdanisur last updated on 07/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{and}\:\:\mathrm{c}\:\:\mathrm{are}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of}\:\:\mathrm{a} \\ $$$$\mathrm{cuboid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{diagonal}\:\boldsymbol{\mathrm{d}}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{d}\:\leqslant\:\sqrt{\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{c}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{a}}} \\ $$ Answered by dumitrel last updated on…