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Question Number 201820 Answers: 1 Comments: 0

((∣3x+1∣−∣x+2∣)/(3−∣2x∣)) ≥ 0 find the solution set.

$$\:\:\:\frac{\mid\mathrm{3x}+\mathrm{1}\mid−\mid\mathrm{x}+\mathrm{2}\mid}{\mathrm{3}−\mid\mathrm{2x}\mid}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}. \\ $$

Question Number 201819 Answers: 0 Comments: 0

If xyz ∈R^+ , xyz=1 , prove that the following inequality holds: (x/(2x^5 +x+4))+(y/(2y^5 +y+4))+(z/(2z^5 +z+4))≥(3/7). Solution please with an advice to get better at inequalities and which book would you recommend. Thanks in advance!

Question Number 201817 Answers: 1 Comments: 0

Question Number 201808 Answers: 1 Comments: 1

Question Number 201806 Answers: 1 Comments: 2

Question Number 201804 Answers: 0 Comments: 0

Question Number 201802 Answers: 0 Comments: 1

Solution of equations like this: ((f(x)))^(1/n) +((g(x)))^(1/n) =((h(x)))^(1/n) If the solution is not obvious we must get rid of the radicals. In the following cases this is easy: (√a)+(√b)=(√c) ⇒ a+2(√(ab))+b=c ⇒ 4ab=(c−a−b)^2 (a)^(1/3) +(b)^(1/3) =(c)^(1/3) ⇒ a+3((ab))^(1/3) ((a)^(1/3) +(b)^(1/3) )+b=c ⇒ a+3((abc))^(1/3) +b=c ⇒ 27abc=(c−a−b)^3 But how to solve for n≥4? I found this formula to get an equation without radicals: a^(1/n) +b^(1/n) =c^(1/n) ⇒ Π_(k=0) ^(n−1) (c−(a^(1/n) +b^(1/n) e^(i((2πk)/n)) )^n ) =0 For n=2, 3 we get above equations. For n=4: c^4 −4(a+b)c^3 +2(3a^2 −62ab+3b^2 )c^2 −4(a+b)(a^2 +30ab+b^2 )c+(a−b)^4 =0 ⇔ 8ab(17c^2 +14(a+b)c+a^2 +b^2 )=(c−a−b)^4 For n=5: 625abc(c^2 +3(a+b)c+a^2 −3ab+b^2 )=(c−a−b)^5 I hope this helps...

$$\mathrm{Solution}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\sqrt[{{n}}]{{f}\left({x}\right)}+\sqrt[{{n}}]{{g}\left({x}\right)}=\sqrt[{{n}}]{{h}\left({x}\right)} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{obvious}\:\mathrm{we}\:\mathrm{must}\:\mathrm{get} \\ $$$$\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radicals}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy}: \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\sqrt{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{2}\sqrt{{ab}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{4}{ab}=\left({c}−{a}−{b}\right)^{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}=\sqrt[{\mathrm{3}}]{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{3}\sqrt[{\mathrm{3}}]{{ab}}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}\right)+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{a}+\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{27}{abc}=\left({c}−{a}−{b}\right)^{\mathrm{3}} \\ $$$$\mathrm{But}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{n}\geqslant\mathrm{4}? \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{get}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{without}\:\mathrm{radicals}: \\ $$$${a}^{\frac{\mathrm{1}}{{n}}} +{b}^{\frac{\mathrm{1}}{{n}}} ={c}^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\:\left({c}−\left({a}^{\frac{\mathrm{1}}{{n}}} +{b}^{\frac{\mathrm{1}}{{n}}} \mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi{k}}{{n}}} \right)^{{n}} \right)\:=\mathrm{0} \\ $$$$\mathrm{For}\:{n}=\mathrm{2},\:\mathrm{3}\:\mathrm{we}\:\mathrm{get}\:\mathrm{above}\:\mathrm{equations}. \\ $$$$\mathrm{For}\:{n}=\mathrm{4}: \\ $$$${c}^{\mathrm{4}} −\mathrm{4}\left({a}+{b}\right){c}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{62}{ab}+\mathrm{3}{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −\mathrm{4}\left({a}+{b}\right)\left({a}^{\mathrm{2}} +\mathrm{30}{ab}+{b}^{\mathrm{2}} \right){c}+\left({a}−{b}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{8}{ab}\left(\mathrm{17}{c}^{\mathrm{2}} +\mathrm{14}\left({a}+{b}\right){c}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\left({c}−{a}−{b}\right)^{\mathrm{4}} \\ $$$$\mathrm{For}\:{n}=\mathrm{5}: \\ $$$$\mathrm{625}{abc}\left({c}^{\mathrm{2}} +\mathrm{3}\left({a}+{b}\right){c}+{a}^{\mathrm{2}} −\mathrm{3}{ab}+{b}^{\mathrm{2}} \right)=\left({c}−{a}−{b}\right)^{\mathrm{5}} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{helps}... \\ $$

Question Number 201798 Answers: 1 Comments: 0

Find the differential of the function: y = (√(x^2 − 1))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\boldsymbol{\mathrm{y}}\:=\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$

Question Number 201796 Answers: 0 Comments: 2

x^4 −15x^2 −30x+104=0 for x∈R x=2, 4 I want to know the best way to arrive at these answers (without guessing). I found one new way. Shall post later.

$${x}^{\mathrm{4}} −\mathrm{15}{x}^{\mathrm{2}} −\mathrm{30}{x}+\mathrm{104}=\mathrm{0} \\ $$$${for}\:{x}\in\mathbb{R}\:\:\:\:\:{x}=\mathrm{2},\:\mathrm{4} \\ $$$${I}\:{want}\:{to}\:{know}\:{the}\:{best}\:{way}\:{to} \\ $$$${arrive}\:{at}\:{these}\:{answers}\:\left({without}\right. \\ $$$$\left.{guessing}\right).\:{I}\:{found}\:{one}\:{new}\:{way}. \\ $$$$\:{Shall}\:{post}\:{later}. \\ $$

Question Number 201793 Answers: 1 Comments: 0

Question Number 201790 Answers: 1 Comments: 1

Question Number 201764 Answers: 1 Comments: 0

1. y = tgx − ctgx → y^′ = ? 2. y = (1 + x^2 ) arctgx → y^′ = ? 3. y = cos^4 x → y^′ = ? 4. { ((x = 2t)),((y = 3t^2 − 5t)) :} → x^′ , y^′ = ?

$$\mathrm{1}.\:\mathrm{y}\:=\:\mathrm{tgx}\:−\:\mathrm{ctgx}\:\:\rightarrow\:\:\mathrm{y}^{'} \:=\:? \\ $$$$\mathrm{2}.\:\mathrm{y}\:=\:\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{arctgx}\:\rightarrow\:\mathrm{y}^{'} \:=\:? \\ $$$$\mathrm{3}.\:\mathrm{y}\:=\:\mathrm{cos}^{\mathrm{4}} \:\mathrm{x}\:\rightarrow\:\mathrm{y}^{'} \:=\:? \\ $$$$\mathrm{4}.\:\begin{cases}{\mathrm{x}\:=\:\mathrm{2t}}\\{\mathrm{y}\:=\:\mathrm{3t}^{\mathrm{2}} \:−\:\mathrm{5t}}\end{cases}\:\:\:\rightarrow\:\:\:\mathrm{x}^{'} \:,\:\mathrm{y}^{'} \:=\:? \\ $$

Question Number 201763 Answers: 5 Comments: 0

Find: 1. ∫ cos3x cosx dx = ? 2. ∫ 3^x sinx dx = ? 3. ∫_(0 ) ^( 1) x e^(−x) dx = ? 4. ∫_1 ^( e) ln^2 x dx = ?

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\int\:\mathrm{cos3x}\:\mathrm{cosx}\:\mathrm{dx}\:=\:? \\ $$$$\mathrm{2}.\:\int\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:\mathrm{sinx}\:\mathrm{dx}\:=\:? \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}\:} ^{\:\mathrm{1}} \:\mathrm{x}\:\mathrm{e}^{−\boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:=\:? \\ $$$$\mathrm{4}.\:\int_{\mathrm{1}} ^{\:\boldsymbol{\mathrm{e}}} \:\mathrm{ln}^{\mathrm{2}} \:\mathrm{x}\:\mathrm{dx}\:=\:? \\ $$

Question Number 201762 Answers: 1 Comments: 0

Question Number 201956 Answers: 1 Comments: 1

Question Number 201727 Answers: 1 Comments: 0

∫_(π/6) ^(π/3) e^(sin x^(cos x^(tan x^(cot x^(sec x^(cosec x) ) ) ) ) ) dx

$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} {e}^{\mathrm{sin}\:{x}^{{c}\mathrm{os}\:{x}^{\mathrm{tan}\:{x}^{\mathrm{cot}\:{x}^{\mathrm{sec}\:{x}^{\mathrm{cosec}\:{x}} } } } } } {dx} \\ $$

Question Number 201729 Answers: 1 Comments: 10

The teacher can choose in 560 ways, provided that there are three students in each team. Knowing that five students do not want to participate, find the number of people willing to participate

Question Number 201724 Answers: 0 Comments: 0

lim_(x→0) ((e^x (x−2)+x+2)/x^3 ) solve it by not using taylor series or l′hopital rule.

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} \left({x}−\mathrm{2}\right)+{x}+\mathrm{2}}{{x}^{\mathrm{3}} }\:{solve}\:{it}\:{by}\:{not}\:{using} \\ $$$${taylor}\:{series}\:{or}\:{l}'{hopital}\:{rule}. \\ $$

Question Number 201722 Answers: 5 Comments: 1