Question Number 143899 by liberty last updated on 19/Jun/21 Answered by ajfour last updated on 19/Jun/21 $${let}\:\:{r}^{\mathrm{10}} ={z} \\ $$$${t}={z}^{\mathrm{4}} −{z}^{\mathrm{3}} +{z}^{\mathrm{2}} −{z}+\mathrm{1} \\ $$$${r}^{\mathrm{5}}…
Question Number 78358 by aliesam last updated on 16/Jan/20 $${Q}.\:{solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}^{{sin}^{\mathrm{2}} \left({x}\right)} }{{sin}^{\mathrm{2}} \left({x}\right)\:}\:+\:\frac{\mathrm{3}^{{cos}^{\mathrm{2}} \left({x}\right)} }{{cos}^{\mathrm{2}} \left({x}\right)}\:=\:\mathrm{6} \\ $$ Commented by MJS…
Question Number 78357 by TawaTawa last updated on 16/Jan/20 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\frac{\sqrt{\mathrm{1}\:+\:\mathrm{6x}}\:\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\mathrm{6x}}}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{3x}}\:\:−\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\:\mathrm{3x}}}}\:\:\:\:\:=\:\:\:\mathrm{4}\:+\:\mathrm{6x} \\ $$$$\mathrm{Ignoring}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$ Commented by MJS last updated on 16/Jan/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{this}\:\mathrm{question} \\ $$$$\mathrm{the}\:\mathrm{left}\:\mathrm{handed}\:\mathrm{side}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for} \\…
Question Number 78353 by TawaTawa last updated on 16/Jan/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 78351 by loveineq. last updated on 16/Jan/20 $$\mathrm{Let}\:\:{a},{b},{c}\:\in\:\mathrm{R}^{+} \:\:\mathrm{and}\:\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{1}\:,\:\mathrm{where}\:\:\mathrm{0}\:<{b}\leqslant\:\mathrm{1}\:. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\frac{\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid}{\mathrm{2}}\:. \\ $$ Commented by loveineq. last updated on 19/Jan/20 $$\mathrm{More}\:\mathrm{stronger}\:\mathrm{is}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid\:. \\ $$…
Question Number 78340 by loveineq. last updated on 16/Jan/20 $$\mathrm{Let}\:\:{a},{b},{c}\:>\:\mathrm{0}\:\:\mathrm{and}\:\:{c}^{\mathrm{2}} \:=\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\:\leqslant\:\mathrm{3}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right)…
Question Number 143875 by ajfour last updated on 19/Jun/21 Commented by ajfour last updated on 19/Jun/21 $${solve}\:{for}\:{m}\:{in}\:{terms}\:{of} \\ $$$${a},{b},{c},{d}. \\ $$ Terms of Service Privacy…
Question Number 78335 by john santu last updated on 16/Jan/20 $${find}\:{the}\:{solution}\:{of} \\ $$$$\frac{{x}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}}\:\geqslant\:\mathrm{0} \\ $$ Answered by MJS last updated on 16/Jan/20 $$\Rightarrow\:{x}>\mathrm{0}\wedge\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}}…
Question Number 12797 by tawa last updated on 01/May/17 $$\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:…\:\mathrm{x}^{\mathrm{49}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{49}} \:−\:\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$ Commented by prakash jain last updated on 01/May/17…
Question Number 78319 by clarissagirl last updated on 16/Jan/20 $$\mathrm{57}+\mathrm{6}{h}=\mathrm{16}{h}−\mathrm{33} \\ $$ Commented by jagoll last updated on 16/Jan/20 $$\mathrm{10}{h}=\mathrm{90}\:\Rightarrow\:{h}=\mathrm{9} \\ $$ Terms of Service…