Question Number 128204 by I want to learn more last updated on 05/Jan/21 $$\mathrm{Let}\:\:\:\mathrm{x},\:\:\mathrm{y},\:\:\mathrm{z}\:\:\:\mathrm{be}\:\mathrm{pairwise}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{numbers},\:\mathrm{if} \\ $$$$\:\:\left(\mathrm{x}\:\:+\:\:\mathrm{y}\:\:−\:\:\mathrm{7}\right)\left[\mathrm{z}\left(\mathrm{x}\:\:+\:\:\mathrm{y}\right)\:\:+\:\:\mathrm{24}\right]\:\:\:=\:\:\left(\mathrm{y}\:\:+\:\:\mathrm{z}\:\:−\:\:\mathrm{7}\right)\left[\mathrm{x}\left(\mathrm{y}\:\:+\:\:\mathrm{z}\right)\:\:+\:\:\mathrm{24}\right]\:\:\:=\:\:\:\left(\mathrm{z}\:\:+\:\:\mathrm{x}\:\:−\:\:\mathrm{7}\right)\left[\mathrm{y}\left(\mathrm{z}\:\:+\:\:\mathrm{x}\right)\:\:+\:\:\mathrm{24}\right] \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{y}^{\mathrm{2}} \:\:+\:\:\mathrm{z}^{\mathrm{2}} \\ $$ Terms of Service…
Question Number 62669 by Sayantan chakraborty last updated on 24/Jun/19 $${calculate}\:{the}\:{value}\:{of}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{1947}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} +\sqrt{\mathrm{2}^{\mathrm{1947}} }} \\ $$ Answered by tanmay last updated on 24/Jun/19 $${a}=\sqrt{\mathrm{2}^{\mathrm{1947}}…
Question Number 128166 by bramlexs22 last updated on 05/Jan/21 $$\:\sqrt{\mathrm{x}−\mathrm{2}}\:+\:\sqrt{\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{4x}+\mathrm{1}}\:=\:\mathrm{10} \\ $$$$\:\mathrm{for}\:\mathrm{x}\in\mathbb{R}\: \\ $$ Answered by liberty last updated on 05/Jan/21 $$\:\sqrt{\mathrm{4x}+\mathrm{1}}\:=\:\left[\:−\sqrt{\mathrm{x}+\mathrm{3}}−\sqrt{\mathrm{x}−\mathrm{2}}+\mathrm{10}\:\right] \\ $$$$\:\mathrm{4x}+\mathrm{1}\:=\:\mathrm{2}\sqrt{\mathrm{x}−\mathrm{2}\:}\:\sqrt{\mathrm{x}+\mathrm{3}}\:−\mathrm{20}\sqrt{\mathrm{x}+\mathrm{3}}\:+\mathrm{2x}\:−\mathrm{20}\sqrt{\mathrm{x}−\mathrm{2}}\:+\mathrm{101} \\…
Question Number 128171 by ajfour last updated on 05/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62623 by meenakshi last updated on 23/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128146 by sahnaz last updated on 04/Jan/21 $$\mathrm{4}<\mathrm{x}<\mathrm{5}\:\mathrm{a}=\sqrt{\mathrm{x}−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{x}−\mathrm{4}}\:\:}\:\:\:\mathrm{b}=\sqrt{\mathrm{x}−\mathrm{4}}−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{x}−\mathrm{a}×\mathrm{b}=? \\ $$ Answered by Fikret last updated on 04/Jan/21 $${a}=\sqrt{{x}−\mathrm{4}}+\mathrm{1} \\ $$$${b}=\sqrt{{x}−\mathrm{4}}−\mathrm{1} \\ $$$${a}.{b}={x}−\mathrm{5} \\…
Question Number 128132 by I want to learn more last updated on 04/Jan/21 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}}\right)\:\:\:=\:\:\:\mathrm{x}^{\mathrm{4}} \:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:,\:\:\:\:\:\mathrm{find}\:\:\mathrm{f}\left(\mathrm{5}\right) \\ $$ Answered by Olaf last updated on 04/Jan/21…
Question Number 62577 by Sr@2004 last updated on 23/Jun/19 Commented by Sr@2004 last updated on 23/Jun/19 $${please}\:{solve}\:\mathrm{12} \\ $$ Answered by som(math1967) last updated on…
Question Number 128090 by ajfour last updated on 04/Jan/21 $${someone}\:{help}\:{cheking}\:{this}: \\ $$$$\:\:\:\:{x}^{\mathrm{3}} ={x}+{c}\:\: \\ $$$${let}\:\:{x}=\left({p}−\mathrm{2}{q}\right)+\left({q}−\mathrm{2}{p}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:−\left({p}+{q}\right) \\ $$$${p}^{\mathrm{3}} −\mathrm{8}{q}^{\mathrm{3}} −\mathrm{6}{pq}\left({p}−\mathrm{2}{q}\right)+ \\ $$$${q}^{\mathrm{3}} −\mathrm{8}{p}^{\mathrm{3}} −\mathrm{6}{pq}\left({q}−\mathrm{2}{p}\right)+…
Question Number 128065 by 0731619177 last updated on 04/Jan/21 Commented by mnjuly1970 last updated on 04/Jan/21 $${lim}_{{n}\rightarrow\infty\:\:\:} \underset{{k}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}}…