Question Number 64508 by lalitchand last updated on 18/Jul/19 $$\mathrm{factorize}\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}+\mathrm{7}\right)+\mathrm{16} \\ $$ Commented by Prithwish sen last updated on 18/Jul/19 $$\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{7}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)+\mathrm{16} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{7}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{15}\right)+\mathrm{16}…
Question Number 130036 by liberty last updated on 22/Jan/21 $$\mathrm{If}\:\begin{cases}{{a}+{b}+{c}\:=\:\mathrm{1}}\\{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:\mathrm{4}}\end{cases} \\ $$$$\:\mathrm{find}\:\frac{\mathrm{1}}{{a}+{bc}}\:+\:\frac{\mathrm{1}}{{b}+{ca}}\:+\:\frac{\mathrm{1}}{{c}+{ab}}. \\ $$ Answered by mnjuly1970 last updated on 22/Jan/21 $${solution}:\:\left\{_{{a}^{\mathrm{3}}…
Question Number 64475 by aliesam last updated on 18/Jul/19 $${solve}\:{the}\:{equation} \\ $$$${sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)+{sin}\left(\mathrm{3}{x}\right)={cos}\left({x}\right)+{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\mathrm{3}{x}\right) \\ $$ Commented by mathmax by abdo last updated on 18/Jul/19 $${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{0}}…
Question Number 130006 by Adel last updated on 21/Jan/21 Answered by Dwaipayan Shikari last updated on 21/Jan/21 $${S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +…. \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +… \\…
Question Number 64447 by Tawa1 last updated on 18/Jul/19 Answered by som(math1967) last updated on 18/Jul/19 $$\left(\mathrm{4}−\mathrm{3}\right)\left(\mathrm{4}+\mathrm{3}\right)\left(\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)…..\left(\mathrm{4}^{\mathrm{32}} +\mathrm{3}^{\mathrm{32}} \right)=\mathrm{4}^{{x}} −\mathrm{3}^{{x}} \:\bigstar \\ $$$$\left(\mathrm{4}^{\mathrm{2}}…
Question Number 129979 by liberty last updated on 21/Jan/21 $$\mathrm{If}\:\begin{cases}{\mid\mathrm{x}\mid\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}}\\{\mathrm{x}\:+\:\mid\mathrm{y}\mid\:−\mathrm{y}\:=\:\mathrm{10}}\end{cases}\:\mathrm{then}\:\mathrm{x}+\mathrm{y}\:? \\ $$ Answered by MJS_new last updated on 21/Jan/21 $$\mid{x}\mid+{x}+{y}=\mathrm{5} \\ $$$${x}\leqslant\mathrm{0}\:\Rightarrow\:{y}=\mathrm{5} \\ $$$$\mathrm{but}\:\mathrm{then}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{eq}.\:\mathrm{gives}\:{x}=\mathrm{10}>\mathrm{0}…
Question Number 64404 by aliesam last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{x}+\mathrm{1}\:=\:\sqrt{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\left(\mathrm{x}+\mathrm{2}\right)}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }}} \\ $$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{4}\right)}}\:}…
Question Number 64378 by aliesam last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{Wow}! \\ $$ Terms of Service Privacy Policy…
Question Number 129841 by Adel last updated on 20/Jan/21 $$\left\{_{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2x}} }+\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{4x}} }+\mathrm{4}^{\mathrm{x}} =\mathrm{5}} ^{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{x}} \:}−\mathrm{2}^{\mathrm{x}} =\mathrm{2}} \right\}\Rightarrow\mathrm{2}^{\mathrm{x}} −\mathrm{8}^{\mathrm{x}} =? \\ $$$$\mathrm{pleas}\:\mathrm{solve}\:\mathrm{thes} \\ $$ Commented by…
Question Number 129832 by ajfour last updated on 19/Jan/21 Commented by ajfour last updated on 19/Jan/21 $${If}\:{p},\:{q},\:{r}\:{are}\:{roots}\:{of}\:{y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$${then}\:{find}\:{p}\:{or}\:{q}\:{or}\:{r},\:{with}\:{the} \\ $$$${help}\:{of}\:{s}. \\ $$ Commented…