Question Number 192855 by pascal889 last updated on 29/May/23 Answered by a.lgnaoui last updated on 29/May/23 $$\begin{cases}{\mathrm{P}^{\mathrm{2}} −\mathrm{2aP}−\mathrm{10P}+\mathrm{2a}^{\mathrm{2}} +\mathrm{6a}\:\:−\mathrm{6}=\mathrm{0}\:\:\left(\mathrm{1}\right)}\\{\mathrm{P}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{27P}\:\:\:\:\:\:\:\:\:+\mathrm{27a}−\mathrm{27}=\mathrm{0}\:\:\left(\mathrm{2}\right)\:}\end{cases} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow\:\mathrm{2a}^{\mathrm{2}} −\mathrm{21a}+\mathrm{21}+\mathrm{P}\left(\mathrm{17}−\mathrm{2a}\right)=\mathrm{0} \\ $$$$…
Question Number 192851 by MATHEMATICSAM last updated on 29/May/23 $$\mathrm{Solve}: \\ $$$$\frac{\mathrm{log}_{{a}^{\mathrm{2}} \sqrt{{x}}} \:{a}}{\mathrm{log}_{\mathrm{2}{x}} \:{a}}\:+\:\mathrm{log}_{{ax}} \:{a}\:.\:\mathrm{log}_{\frac{\mathrm{1}}{{a}}} \:\mathrm{2}{x}\:=\:\mathrm{0} \\ $$ Commented by Frix last updated on…
Question Number 127306 by Study last updated on 28/Dec/20 Commented by Study last updated on 28/Dec/20 $${help}\:{me} \\ $$ Commented by mr W last updated…
Question Number 192839 by Mingma last updated on 29/May/23 Answered by witcher3 last updated on 02/Jun/23 $$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{2},\mathrm{7}\:\mathrm{can}'\mathrm{t}\:\mathrm{bee}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{and}\:\mathrm{11}\mid\mathrm{n}\: \\ $$$$\mathrm{if}\:\mathrm{n}≢\mathrm{0}\left[\mathrm{11}\right] \\ $$$$\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{4}^{\mathrm{n}} +\mathrm{7}^{\mathrm{n}}…
Question Number 127301 by Study last updated on 28/Dec/20 Commented by Study last updated on 28/Dec/20 $${what}\:{price}\:{will}\:{scales}\:{read}???? \\ $$ Commented by mr W last updated…
Question Number 192811 by Abdullahrussell last updated on 28/May/23 Answered by AST last updated on 28/May/23 $$\Sigma\frac{\mathrm{1}}{{x}+{yz}}=\Sigma\frac{{x}}{{x}^{\mathrm{2}} +{xyz}}=\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{5}}+\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{5}}+\frac{{z}}{{z}^{\mathrm{2}} +\mathrm{5}} \\ $$$$=\frac{\Sigma\left({xy}^{\mathrm{2}} +\mathrm{5}{x}\right)\left({z}^{\mathrm{2}} +\mathrm{5}\right)=\Sigma\left({xy}^{\mathrm{2}}…
Question Number 192793 by York12 last updated on 27/May/23 $$\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{1}=\boldsymbol{{y}}^{\mathrm{2}} \:\boldsymbol{{where}}\:\boldsymbol{{y}}\:\boldsymbol{{is}}\:\boldsymbol{{positive}}\:\boldsymbol{{integer}}\:\boldsymbol{{number}} \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{positive}}\:\boldsymbol{{integal}}\:\boldsymbol{{values}}\:\boldsymbol{{of}}\:\left(\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{which}}\:\boldsymbol{{that}}\:\boldsymbol{{holds}} \\ $$ Answered by AST last updated…
Question Number 192786 by MM42 last updated on 27/May/23 $${N}=<{aabb}>\in\mathbb{N}\:\:\&\:\:{N}\:\:{is}\:\:{perfect}\:{square} \\ $$$${find}\:\:{N}\:\:? \\ $$$$ \\ $$ Answered by AST last updated on 27/May/23 $$\mathrm{1100}{a}+\mathrm{11}{b}={x}^{\mathrm{2}} \Rightarrow\mathrm{11}\left(\mathrm{100}{a}+{b}\right)={x}^{\mathrm{2}}…
Question Number 192780 by ajfour last updated on 27/May/23 Commented by ajfour last updated on 27/May/23 Find m in terms of c, hence p. Answered by ajfour last updated on 27/May/23 $${y}={x}^{\mathrm{3}}…
Question Number 192777 by York12 last updated on 26/May/23 $$\sqrt{{x}^{\mathrm{2}} +{ax}−\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +{bx}−\mathrm{1}}=\sqrt{{a}}−\sqrt{{b}} \\ $$$${find}\:{x}\:. \\ $$ Commented by Frix last updated on 27/May/23 $$\mathrm{Obviously}\:{x}=\mathrm{1} \\…