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AlgebraQuestion and Answers: Page 65

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remark to question 195301 and similar ones x^2 +y=a x+y^2 =b a, b >0 how many solutions depending on a, b?

$$\mathrm{remark}\:\mathrm{to}\:\mathrm{question}\:\mathrm{195301}\:\mathrm{and}\:\mathrm{similar}\:\mathrm{ones} \\ $$$${x}^{\mathrm{2}} +{y}={a} \\ $$$${x}+{y}^{\mathrm{2}} ={b} \\ $$$${a},\:{b}\:>\mathrm{0} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{depending}\:\mathrm{on}\:{a},\:{b}? \\ $$

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it is given a,b,c ∈ N^∗ and ab<c . Prove that a+b≤c.

$$\:\:{it}\:{is}\:{given}\:{a},{b},{c}\:\in\:\mathbb{N}^{\ast} \:\:{and}\:\:{ab}<{c}\:.\:{Prove}\:{that}\:{a}+{b}\leqslant{c}. \\ $$

Question Number 195288 Answers: 0 Comments: 1

x, y, z∈R_+ , P = (x/(x + y)) + (y/(y + z)) + (z/(z + x)), Q = (y/(x + y)) + (z/(y + z)) + (x/(z + x)), Q = (z/(x + y)) + (x/(y + z)) + (y/(z + x)). f(x, y, z)=max{P, Q, R}, find f_(min) .

$${x},\:{y},\:{z}\in\mathbb{R}_{+} , \\ $$$${P}\:=\:\frac{{x}}{{x}\:+\:{y}}\:+\:\frac{{y}}{{y}\:+\:{z}}\:+\:\frac{{z}}{{z}\:+\:{x}}, \\ $$$${Q}\:=\:\frac{{y}}{{x}\:+\:{y}}\:+\:\frac{{z}}{{y}\:+\:{z}}\:+\:\frac{{x}}{{z}\:+\:{x}}, \\ $$$${Q}\:=\:\frac{{z}}{{x}\:+\:{y}}\:+\:\frac{{x}}{{y}\:+\:{z}}\:+\:\frac{{y}}{{z}\:+\:{x}}. \\ $$$${f}\left({x},\:{y},\:{z}\right)=\mathrm{max}\left\{{P},\:{Q},\:{R}\right\},\:\mathrm{find}\:{f}_{\mathrm{min}} . \\ $$

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∫(dx/(cos^3 x(√(4sin xcos x))))

$$\int\frac{{dx}}{\mathrm{cos}\:^{\mathrm{3}} {x}\sqrt{\mathrm{4sin}\:{x}\mathrm{cos}\:{x}}} \\ $$

Question Number 195254 Answers: 1 Comments: 0

∫^(spillover) ((sin^2 xcos^2 x)/((sin^5 x+cos^3 xsin^2 x+sin^3 xcos^2 x+cos^5 x)^2 ))dx

$$\int^{\boldsymbol{{spillover}}} \frac{\mathrm{sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}{\left(\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{5}} {x}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 195253 Answers: 2 Comments: 1

If 10sin^4 x+15cos^4 x=6. find the value of 27cosec^6 x+8sec^6 x

$${If}\:\mathrm{10sin}\:^{\mathrm{4}} {x}+\mathrm{15cos}\:^{\mathrm{4}} {x}=\mathrm{6}. \\ $$$${find}\:{the}\:{value}\:{of} \\ $$$$\mathrm{27cosec}\:^{\mathrm{6}} {x}+\mathrm{8sec}\:^{\mathrm{6}} {x} \\ $$$$ \\ $$

Question Number 195252 Answers: 1 Comments: 0

∫_(spillover) (dx/( (√e^(5x) ) (√((e^(2x) +e^(−2x) )^3 ))))

$$\int_{\boldsymbol{{spillover}}} \:\:\:\:\:\:\frac{{dx}}{\:\sqrt{{e}^{\mathrm{5}{x}} }\:\sqrt{\left({e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} \right)^{\mathrm{3}} }} \\ $$

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If x^([16(log _5 x)^3 −68log _5 x]) =5^(−16) then Find the the product of x

$${If}\:\:{x}^{\left[\mathrm{16}\left(\mathrm{log}\:_{\mathrm{5}} {x}\right)^{\mathrm{3}} −\mathrm{68log}\:_{\mathrm{5}} {x}\right]} =\mathrm{5}^{−\mathrm{16}} \: \\ $$$$\:{then}\:{Find}\:{the}\:{the}\:{product}\:{of}\:{x} \\ $$$$ \\ $$

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below equestion is show elips and hypharabollah (x^2 /(cos3))+(y^2 /(sin3))=1

$${below}\:{equestion}\:{is}\:{show}\:\:{elips}\:{and} \\ $$$${hypharabollah} \\ $$$$\frac{{x}^{\mathrm{2}} }{{cos}\mathrm{3}}+\frac{{y}^{\mathrm{2}} }{{sin}\mathrm{3}}=\mathrm{1} \\ $$

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α_1 ^3 [((Π_(i=2) ^n (x−α_i ))/(Π_(i=2) ^n (α_1 −α_i )))]+Σ_(j=2) ^n (α_j ^3 [((Π_(i=1) ^(j−1) (x−α_i )Π_(i=j+1) ^n (x−α_j ))/(Π_(i=1) ^(j−1) (α_j −α_i )Π_(i=j+1) ^n (α_j −α_i )))]+α_n ^3 [((Π_(i=1) ^(n−1) (x−α_i ))/(Π_(i=1) ^(n−1) (α_n −α_i )))]−x^3 =0 solve for x . [ where n≥5 ]

$$ \\ $$$$\alpha_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\alpha_{\mathrm{1}} −\alpha_{{i}} \right)}\right]+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\alpha_{{j}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{j}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)}\right]+\alpha_{{n}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\alpha_{{n}} −\alpha_{{i}} \right)}\right]−{x}^{\mathrm{3}} =\mathrm{0}\right. \\ $$$${solve}\:{for}\:{x}\:.\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{where}\:{n}\geqslant\mathrm{5}\:\right] \\ $$

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