Question Number 150968 by mathdanisur last updated on 17/Aug/21 Answered by dumitrel last updated on 17/Aug/21 $$\Sigma\frac{\left(\mathrm{2}{x}+{y}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +{xy}+\mathrm{2}{y}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\leqslant\Sigma\mathrm{3}\left({x}+{y}\right)=\mathrm{6}\left({x}+{y}+{z}\right) \\ $$ Commented by…
Question Number 19900 by Tinkutara last updated on 17/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{identity}\:\mathrm{in}\:{x}: \\ $$$$\frac{\left({x}−{a}\right)\left({x}−{b}\right)}{\left({c}−{a}\right)\left({c}−{b}\right)}+\frac{\left({x}−{b}\right)\left({x}−{c}\right)}{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{\left({x}−{c}\right)\left({x}−{a}\right)}{\left({b}−{c}\right)\left({b}−{a}\right)}=\mathrm{1} \\ $$ Answered by Rasheed.Sindhi last updated on 17/Aug/17 $$−\frac{\left({x}−{a}\right)\left({x}−{b}\right)}{\left({c}−{a}\right)\left(\mathrm{b}−\mathrm{c}\right)}−\frac{\left({x}−{b}\right)\left({x}−{c}\right)}{\left({a}−{b}\right)\left(\mathrm{c}−\mathrm{a}\right)}−\frac{\left({x}−{c}\right)\left({x}−{a}\right)}{\left({b}−{c}\right)\left(\mathrm{a}−\mathrm{b}\right)}=\mathrm{1} \\ $$$$ \\…
Question Number 19898 by Tinkutara last updated on 17/Aug/17 $$\mathrm{Simplify}\::\:{i}\:\mathrm{log}\:\left(\frac{{x}\:−\:{i}}{{x}\:+\:{i}}\right). \\ $$ Answered by ajfour last updated on 22/Aug/17 $${let}\:{t}={i}\mathrm{ln}\:\left(\frac{{x}−{i}}{{x}+{i}}\right) \\ $$$${and}\:{let}\:{z}=\frac{{x}−{i}}{{x}+{i}}\: \\ $$$$\Rightarrow\:\mid{z}\mid=\mathrm{1} \\…
Question Number 150966 by mathdanisur last updated on 17/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\in\mathbb{N}^{+} \:\:\mathrm{then}\:\mathrm{determine}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{prime}\:\mathrm{numbers}\:\boldsymbol{\mathrm{p}}\:\mathrm{which}\:\mathrm{satisfy} \\ $$$$\left(\mathrm{p}\:+\:\mathrm{2}\right)^{\boldsymbol{\mathrm{a}}} \:=\:\left(\mathrm{p}\:-\:\mathrm{2}\right)^{\boldsymbol{\mathrm{b}}} \\ $$ Commented by Rasheed.Sindhi last updated on 17/Aug/21…
Question Number 150960 by john_santu last updated on 17/Aug/21 $$\mathrm{given}\:\mathrm{x}\:,\mathrm{y}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\: \\ $$$$\mathrm{with}\:\mathrm{x}<\mathrm{y}\:\mathrm{and}\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{20l8}\:=\mathrm{30y}^{\mathrm{2}} −\mathrm{300y}+\mathrm{30l8} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$ Commented by otchereabdullai@gmail.com last updated on…
Question Number 85412 by Power last updated on 21/Mar/20 Commented by jagoll last updated on 22/Mar/20 $$\mathrm{by}\:\mathrm{observe} \\ $$$$\mathrm{x}\:=\:\mathrm{y} \\ $$$$\left(\mathrm{2x}\right)^{\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{3}} \:\Rightarrow\:\mathrm{y}\:=\:\mathrm{x}=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{2x}\right)^{\mathrm{2}}…
Question Number 85418 by oustmuchiya@gmail.com last updated on 21/Mar/20 $${Find}\:{the}\:{term}\:{indepent}\:{of}\:{x}\:{in}\:{the}\:{expression}\:{of}\:\left(\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{9}} \\ $$ Commented by jagoll last updated on 22/Mar/20 $$\mathrm{let}\:\mathrm{2x}=\:\mathrm{u} \\ $$$$\left(\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}\right)^{\mathrm{3}} =\:\mathrm{u}^{\mathrm{3}} −\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{\mathrm{u}}\right)+\mathrm{3u}\left(\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}}…
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Question Number 19875 by ajfour last updated on 16/Aug/17 Commented by ajfour last updated on 17/Aug/17 $${Find}\:{z}_{\mathrm{0}} \:,\:\frac{{p}}{{p}+{q}}\:,\:\frac{{r}}{{r}+{s}}\:;\:{in}\:{terms}\:{of} \\ $$$$\:\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{and}\:{z}_{\mathrm{4}} \:. \\…
Question Number 150918 by mathdanisur last updated on 16/Aug/21 Commented by puissant last updated on 16/Aug/21 $$=\:\left[\frac{\underset{{k}=\mathrm{0}} {\overset{\mathrm{1995}} {\sum}}\mathrm{2}^{{k}} }{\mathrm{1997}}\right]\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1997}}\left(\frac{\mathrm{1}−\mathrm{2}^{\mathrm{1996}} }{\mathrm{1}−\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1997}}\left(\mathrm{2}^{\mathrm{1996}}…