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AlgebraQuestion and Answers: Page 57

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 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 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 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 201722 Answers: 5 Comments: 1

Question Number 201715 Answers: 0 Comments: 0

Use Cauchy Riemann to verify whether f(z)=(1/(z(z+1))) is analytic.

$${Use}\:{Cauchy}\:{Riemann}\:{to}\:{verify}\:{whether}\: \\ $$$${f}\left({z}\right)=\frac{\mathrm{1}}{{z}\left({z}+\mathrm{1}\right)}\:{is}\:{analytic}. \\ $$

Question Number 201728 Answers: 1 Comments: 0

cos^2 4x ∙ sin^2 4x = 0,25 for equation [0;90] how many roots are there in the piece?

$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{4x}\:\centerdot\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{4x}\:=\:\mathrm{0},\mathrm{25}\:\mathrm{for}\:\mathrm{equation} \\ $$$$\left[\mathrm{0};\mathrm{90}\right]\:\mathrm{how}\:\mathrm{many}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{there}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{piece}? \\ $$

Question Number 201702 Answers: 1 Comments: 1

Find: ∫_1 ^( 3) dx ∫_x ^( x^3 ) (x − y) dy

$$\mathrm{Find}: \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{3}} \:\mathrm{dx}\:\int_{\boldsymbol{\mathrm{x}}} ^{\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} } \:\left(\mathrm{x}\:−\:\mathrm{y}\right)\:\mathrm{dy} \\ $$

Question Number 201689 Answers: 1 Comments: 0

Question Number 201679 Answers: 5 Comments: 0

Question Number 201629 Answers: 0 Comments: 3

Question Number 201615 Answers: 1 Comments: 0

x,y,z ∈ R a,b,c>0 prove that: (x^2 /a) + (y^2 /b) + (z^2 /c) ≥ (((x + y + z)^2 )/(a + b + c))

$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{c}}\:\geqslant\:\frac{\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)^{\mathrm{2}} }{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}} \\ $$

Question Number 201613 Answers: 2 Comments: 0

x,y,z ∈ R { ((xy + yz + zx = 3)),((x + y + z = 5)) :} → max(z) = ?

$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}\:=\:\mathrm{3}}\\{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{5}}\end{cases}\:\:\:\:\:\rightarrow\:\:\:\:\mathrm{max}\left(\boldsymbol{\mathrm{z}}\right)\:=\:? \\ $$

Question Number 201547 Answers: 1 Comments: 0

Question Number 201557 Answers: 2 Comments: 0

5 ∙ 555...5_( 50) find the sum of the digits of the product.

$$\mathrm{5}\:\centerdot\:\underset{\:\mathrm{50}} {\underbrace{\mathrm{555}...\mathrm{5}}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{product}. \\ $$

Question Number 201477 Answers: 1 Comments: 0

how to prove that (3d_3 +4d_2 +3d_1 )^2 ≤5(d_1 ^2 +d_2 ^2 +d_3 ^2 +(d_2 +d_1 )^2 +(d_3 +d_2 )^2 +(d_1 +d_2 +d_3 )^2 )

Question Number 201464 Answers: 2 Comments: 0

if f(2) = 3 and f(4) = 5 find ∫_2 ^( 4) f(x) ∙ f^′ (x) dx = ?

$$\mathrm{if}\:\:\:\mathrm{f}\left(\mathrm{2}\right)\:=\:\mathrm{3}\:\:\:\mathrm{and}\:\:\:\mathrm{f}\left(\mathrm{4}\right)\:=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\:\int_{\mathrm{2}} ^{\:\mathrm{4}} \:\mathrm{f}\left(\mathrm{x}\right)\:\centerdot\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$

Question Number 201441 Answers: 0 Comments: 0

Let f(x) and g(x) be given by f(x)= (1/x) +(1/(x−2)) +(1/(x−4)) + ... +(1/(x−2018)) and g(x)=(1/(x−1)) +(1/(x−3)) +(1/(x−5)) +...+ (1/(x−2017)). Prove that ∣ f(x)−g(x)∣ >2 for any non−integer real number x satisfying 0<x<2018.

$${Let}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{be}\:{given}\:{by}\: \\ $$$$\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}−\mathrm{2}}\:+\frac{\mathrm{1}}{{x}−\mathrm{4}}\:+\:...\:+\frac{\mathrm{1}}{{x}−\mathrm{2018}} \\ $$$$\:{and}\: \\ $$$$\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{{x}−\mathrm{3}}\:+\frac{\mathrm{1}}{{x}−\mathrm{5}}\:+...+\:\frac{\mathrm{1}}{{x}−\mathrm{2017}}. \\ $$$$\:\:{Prove}\:{that}\:\:\mid\:{f}\left({x}\right)−{g}\left({x}\right)\mid\:>\mathrm{2} \\ $$$$\:\:{for}\:{any}\:{non}−{integer}\:{real}\:{number} \\ $$$$\:\:{x}\:{satisfying}\:\mathrm{0}<{x}<\mathrm{2018}.\: \\ $$

Question Number 201430 Answers: 1 Comments: 0

Find: (2/(35)) + (2/(63)) + (2/(99)) + (2/(143)) = ?

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{2}}{\mathrm{35}}\:+\:\frac{\mathrm{2}}{\mathrm{63}}\:+\:\frac{\mathrm{2}}{\mathrm{99}}\:+\:\frac{\mathrm{2}}{\mathrm{143}}\:=\:? \\ $$

Question Number 201408 Answers: 3 Comments: 0

Question Number 201388 Answers: 2 Comments: 0

Question Number 201343 Answers: 1 Comments: 0

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