Question Number 20051 by Tinkutara last updated on 22/Aug/17 $$\mathrm{If}\:{x}\:\in\:{R}\:\mathrm{then}\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:{a}}{{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{3}{a}}\:\mathrm{can}\:\mathrm{take}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{values}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:{a}\:\in\:\left(\mathrm{0},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:{a}\:\in\:\left[\mathrm{0},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{3}\right)\:{a}\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{4}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$ Answered…
Question Number 20047 by tawa tawa last updated on 20/Aug/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{x}}\:+\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$ Answered by ajfour last updated on 21/Jun/18 $${first}\:{term}\:{is}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}}}\:\:{or}\:\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}}\:\:? \\ $$$${assuming}\:{it}\:{is}\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}}\:;…
Question Number 151115 by mathdanisur last updated on 18/Aug/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{xy}\:+\:\mathrm{1}\:=\:\mathrm{0}}\end{cases} \\ $$ Answered by dumitrel last updated on 18/Aug/21…
Question Number 151104 by john_santu last updated on 18/Aug/21 Commented by john_santu last updated on 18/Aug/21 find the center of gravity with respect to point O Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\bullet\:\mathrm{vertical}…
Question Number 151101 by mathdanisur last updated on 18/Aug/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left[\frac{\mathrm{x}}{\mathrm{2}}\right]\:+\:\left[\frac{\mathrm{3x}}{\mathrm{5}}\right]\:=\:\left[\frac{\mathrm{x}}{\mathrm{10}}\right]\:+\:\mathrm{x}\:,\:\:\mathrm{where}\:\mathrm{we} \\ $$$$\mathrm{denoting}\:\mathrm{by}\:\left[\boldsymbol{\mathrm{x}}\right]\:\mathrm{the}\:\mathrm{great}\:\mathrm{integer}\:\mathrm{part} \\ $$$$\mathrm{of}\:\boldsymbol{\mathrm{x}}. \\ $$ Answered by dumitrel last updated on 18/Aug/21…
Question Number 151100 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mathrm{0}\leqslant\mathrm{x};\mathrm{y};\mathrm{z}\leqslant\mathrm{k}\:\:\mathrm{and}\:\:\mathrm{k}>\mathrm{0}\:\:\mathrm{then}: \\ $$$$\mathrm{y}\left(\mathrm{x}\:-\:\mathrm{z}\right)\:-\:\mathrm{z}\left(\mathrm{x}\:-\:\mathrm{k}\right)\:\leqslant\:\mathrm{k}^{\mathrm{2}} \\ $$ Answered by dumitrel last updated on 18/Aug/21 $$\Leftrightarrow{y}\left({x}−{z}\right)+{z}\left({k}−{x}\right)\leqslant{k}^{\mathrm{2}} \\ $$$${I}.\:\:{If}\:{x}\leqslant{z}\Rightarrow{y}\left({x}−{z}\right)\leqslant\mathrm{0} \\…
Question Number 85554 by luxlavanish last updated on 22/Mar/20 $$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\mathrm{0} \\ $$ Commented by naka3546 last updated on 23/Mar/20 $${what}'{s}\:\:{the}\:\:{question}\:? \\ $$ Terms of…
Question Number 151085 by mathdanisur last updated on 18/Aug/21 Answered by talminator2856791 last updated on 18/Aug/21 $$\: \\ $$$$\:\frac{{a}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\left({a}+\mathrm{3}\right)^{\mathrm{2}} }{\left({a}+\mathrm{4}\right)^{\mathrm{2}} −\left({a}+\mathrm{3}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}\right)^{\mathrm{2}}…
Question Number 151081 by malwan last updated on 18/Aug/21 $${if}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${and}\:{f}\left({x}+\mathrm{3}\right)=\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\:{find} \\ $$$$\left.\mathrm{1}\left.\right)\:{a}+{b}+{c}\:\:\:\:\:\mathrm{2}\right)\:{a}−{b}+{c} \\ $$ Answered by mr W last updated on…
Question Number 20013 by Tinkutara last updated on 20/Aug/17 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\frac{\mathrm{1}}{{x}\:−\:{a}}\:+\:\frac{\mathrm{1}}{{x}\:−\:{b}} \\ $$$$+\:\frac{\mathrm{1}}{{x}\:−\:{c}}\:=\:\mathrm{0}\:\mathrm{can}\:\mathrm{have}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{equal} \\ $$$$\mathrm{roots}\:\mathrm{if}\:{a}\:=\:{b}\:=\:{c}. \\ $$ Commented by ajfour last updated on 20/Aug/17 $${how}\:{did}\:{u}\:{solve}\:{it}\:,\:{please}\:{let}\:{me} \\…