Question Number 200979 by sonukgindia last updated on 27/Nov/23 Answered by Frix last updated on 27/Nov/23 $$\mathrm{Qu}\:\mathrm{200934} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 201004 by Mingma last updated on 27/Nov/23 Answered by ajfour last updated on 27/Nov/23 Commented by Mingma last updated on 27/Nov/23 Very elegant, sir! Answered…
Question Number 200972 by Mingma last updated on 27/Nov/23 Answered by AST last updated on 27/Nov/23 $${a}^{\mathrm{2}} ={b}^{\mathrm{2}} \left({b}+\mathrm{2}\right);{a}={b}\sqrt{{b}+\mathrm{2}} \\ $$$${b}=\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}\Rightarrow\left({a},{b}\right)=\left(\frac{{x}^{\mathrm{3}} −\mathrm{2}{xy}^{\mathrm{2}} }{{y}^{\mathrm{3}} },\frac{{x}^{\mathrm{2}}…
Question Number 200973 by Mingma last updated on 27/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201000 by mokys last updated on 27/Nov/23 $${find}\: \\ $$$$ \\ $$$$\left.\:\mathrm{1}\right)\left(\mathrm{0},\mathrm{2}\right)\:\cup\:\left\{\mathrm{3}\right\}\:\: \\ $$$$ \\ $$$$\left.\:\:\mathrm{2}\right)\left[\mathrm{0},\mathrm{2}\right]\:\cup\:\left\{\mathrm{3}\right\} \\ $$$$ \\ $$$$\left.\:\:\mathrm{3}\right)\:\left(−\mathrm{5},\mathrm{5}\right)\:\cup\:\left\{\mathrm{6}\right\} \\ $$$$\:\: \\…
Question Number 200968 by Mastermind last updated on 27/Nov/23 Commented by AST last updated on 27/Nov/23 $${P}\overset{\mathrm{10}} {−}{S}\overset{\mathrm{12}} {−}{U}\overset{\mathrm{8}} {−}{T}\overset{\mathrm{7}} {−}{R}\overset{\mathrm{15}} {−}{Q}\overset{\mathrm{11}} {−}{P}\Rightarrow{min}=\mathrm{63} \\ $$…
Question Number 200970 by Mastermind last updated on 27/Nov/23 Answered by JDamian last updated on 27/Nov/23 $${Really}? \\ $$ Answered by mr W last updated…
Question Number 200971 by Mingma last updated on 27/Nov/23 Answered by AST last updated on 27/Nov/23 $${WLOG},{let}\:{a}\:{be}\:{the}\:{max}\:{element} \\ $$$${abcd}={a}+{b}+{c}+{d}\leqslant\mathrm{4}{a}\Rightarrow{bcd}\leqslant\mathrm{4} \\ $$$${bcd}=\mathrm{1}\Rightarrow{b}={c}={d}=\mathrm{1}\Rightarrow{a}+\mathrm{3}={a}\left({absurd}\right) \\ $$$${bcd}=\mathrm{2}\Rightarrow{b}+{c}+{d}=\mathrm{4}\Rightarrow\mathrm{4}+{a}=\mathrm{2}{a}\Rightarrow{a}=\mathrm{4} \\ $$$$\Rightarrow\left({a},{b},{c},{d}\right)=\left(\mathrm{4},\mathrm{2},\mathrm{1},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{2},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{1},\mathrm{2}\right)…
Question Number 200964 by Mastermind last updated on 27/Nov/23 Answered by MM42 last updated on 27/Nov/23 $$\mathrm{3} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 200960 by sonukgindia last updated on 27/Nov/23 Answered by Frix last updated on 27/Nov/23 $${t}=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\int…=\int\left(\mathrm{1}−\frac{\mathrm{1}}{{t}}\right){dt}={t}−\mathrm{ln}\:{t}\:= \\ $$$$=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}}…