Question Number 78694 by TawaTawa last updated on 19/Jan/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:−\:\left(\mathrm{y}\:−\:\mathrm{z}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{10}\:\:\:\:\:\:…\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\mathrm{y}^{\mathrm{2}} \:−\:\left(\mathrm{z}\:−\:\mathrm{x}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{5}\:\:\:\:\:\:…\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{x}\:\:−\:\mathrm{y}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{2}\:\:\:\:\:\:…\:\left(\mathrm{iii}\right) \\ $$ Commented…
Question Number 13155 by Tinkutara last updated on 15/May/17 Answered by mrW1 last updated on 16/May/17 $$=\left(\frac{{x}+{y}}{{x}}\right)\left(\frac{{y}+{z}}{{y}}\right)\left(\frac{{z}+{x}}{{z}}\right) \\ $$$$=\left(\mathrm{1}+\frac{{y}}{{x}}\right)\left(\mathrm{1}+\frac{{z}}{{y}}\right)\left(\mathrm{1}+\frac{{x}}{{z}}\right) \\ $$$$\geqslant\mathrm{2}\sqrt{\frac{{y}}{{x}}}×\mathrm{2}\sqrt{\frac{{z}}{{y}}}×\mathrm{2}\sqrt{\frac{{x}}{{z}}}=\mathrm{8} \\ $$$$ \\ $$$${x}+{y}+{z}=\mathrm{1}\:{is}\:{not}\:{necessary}!…
Question Number 13153 by Tinkutara last updated on 15/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17 $$\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{d}\right)= \\ $$$$\left({abcd}+{a}\right)\left({abcd}+{b}\right)\left({abcd}+{c}\right)\left({abcd}+{d}\right)= \\ $$$${abcd}\left({dbc}+\mathrm{1}\right)\left({acd}+\mathrm{1}\right)\left({abd}+\mathrm{1}\right)\left({abc}+\mathrm{1}\right)= \\ $$$$\left({abc}+\mathrm{1}\right)\left({abd}+\mathrm{1}\right)\left({acd}+\mathrm{1}\right)\left({dbc}+\mathrm{1}\right)\geqslant \\ $$$$\geqslant\mathrm{2}\sqrt{{abc}×\mathrm{1}}×\mathrm{2}\sqrt{{abd}×\mathrm{1}}×\mathrm{2}\sqrt{{acd}×\mathrm{1}}×\mathrm{2}\sqrt{{dbc}×\mathrm{1}}=…
Question Number 13145 by Joel577 last updated on 15/May/17 $$\sqrt[{\sqrt[{\sqrt{\mathrm{5}}}]{\mathrm{3}}}]{\mathrm{64}}\:=\:{x} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{write}\:{x}\:\mathrm{into}\:\mathrm{fraction}\:\mathrm{exponent}\:\mathrm{form}? \\ $$ Answered by mrW1 last updated on 16/May/17 $${x}=\mathrm{64}^{\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}}} }} }} \\…
Question Number 144196 by qaz last updated on 23/Jun/21 $$\mathrm{Let}\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\mathrm{and}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3}.\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{c}^{\mathrm{2}} \right)\leqslant\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{abc}}}\right)^{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144199 by qaz last updated on 23/Jun/21 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{6}+…}}}}}=\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}×\mathrm{2}}{\mathrm{1}+\frac{\mathrm{2}×\mathrm{3}}{\mathrm{1}+\frac{\mathrm{3}×\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}×\mathrm{5}}{\mathrm{1}+\frac{\mathrm{5}×\mathrm{6}}{\mathrm{1}+…}}}}}} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 78650 by Pratah last updated on 19/Jan/20 Commented by mr W last updated on 19/Jan/20 $$={x}^{\mathrm{2}} +\mathrm{1} \\ $$ Commented by john santu…
Question Number 144190 by mathdanisur last updated on 22/Jun/21 $$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{5}{n}}{{n}^{\mathrm{2}} \:+\:\mathrm{3}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 23/Jun/21 $$\mathrm{this}\:\mathrm{serie}\:\mathrm{is}\:\mathrm{divergent}\:\mathrm{due}\:\mathrm{to}\:\frac{\mathrm{5n}}{\mathrm{n}^{\mathrm{2}}…
Question Number 144177 by mathdanisur last updated on 22/Jun/21 Answered by mindispower last updated on 22/Jun/21 $${diverge}\: \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$\mathrm{1}−{cos}\left({x}\right)+{x}^{\mathrm{2}} \sim\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}}…
Question Number 144170 by mathdanisur last updated on 22/Jun/21 Answered by Olaf_Thorendsen last updated on 22/Jun/21 $$\mathrm{P}\left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}{a}_{{k}} {x}^{{k}} \\ $$$$\mathrm{P}\left({m}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}{a}_{{k}} {m}^{{k}}…