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Question Number 202536    Answers: 2   Comments: 0

$$\:\:\: \\ $$

Question Number 202535    Answers: 1   Comments: 0

Solve for x ((x^3 +3x−3))^(1/3) +((−x^3 −3x+5))^(1/3) =2 (An alteration of Q#202500)

$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$

Question Number 202532    Answers: 0   Comments: 1

please help

$${please}\:{help} \\ $$

Question Number 202530    Answers: 1   Comments: 0

Question Number 202522    Answers: 0   Comments: 0

Question Number 202514    Answers: 0   Comments: 0

Question Number 202511    Answers: 0   Comments: 0

Question Number 202500    Answers: 4   Comments: 0

Question Number 202497    Answers: 3   Comments: 0

If the difference of the roots of x^2 + 2px + q = 0 is equal to the difference of the roots of x^2 + 2qx + p = 0 [p ≠ q] then show that p + q + 1 = 0.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$

Question Number 202490    Answers: 2   Comments: 2

Question Number 202485    Answers: 0   Comments: 3

Question Number 202477    Answers: 3   Comments: 0

If x = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) then show that b^2 x^2 − 2a^2 x + b^2 = 0.

$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0}. \\ $$

Question Number 202473    Answers: 0   Comments: 0

Find: 1. Σ_(n=1) ^∞ ((2^n ∙ n!)/((2n)!)) 2. Σ_(n=1) ^∞ (((x + 5)^n )/(3^(n+1) ∙ n ∙ ln(n)))

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:\centerdot\:\mathrm{n}!}{\left(\mathrm{2n}\right)!}\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{5}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\centerdot\:\mathrm{n}\:\centerdot\:\mathrm{ln}\left(\mathrm{n}\right)} \\ $$

Question Number 202468    Answers: 1   Comments: 0

Find: 1. Σ_(n=1) ^( ∞) ((16)/(16n^2 − 8n − 3)) = ? 2. Σ_(n=1) ^( ∞) (((−1)^n )/(2n^3 )) = ?

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\mathrm{16}}{\mathrm{16n}^{\mathrm{2}} \:−\:\mathrm{8n}\:−\:\mathrm{3}}\:=\:?\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{2n}^{\mathrm{3}} }\:=\:? \\ $$

Question Number 202466    Answers: 1   Comments: 0

lim_(x→+∞) ((x(√(ln (x^2 +1))))/(1+e^(x−3) ))

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{1}+{e}^{{x}−\mathrm{3}} } \\ $$

Question Number 202464    Answers: 0   Comments: 0

Question Number 202462    Answers: 1   Comments: 0

Question Number 202459    Answers: 3   Comments: 0

If the difference of two roots of x^2 − lx + m = 0 is 1 then prove that l^2 + 4m^2 = (1 + 2m)^2 .

$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:{lx}\:+\:{m}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \:. \\ $$

Question Number 202449    Answers: 0   Comments: 0

Question Number 202448    Answers: 2   Comments: 0

Question Number 202447    Answers: 2   Comments: 0

rationnalise le denominateur de x = (((2)^(1/(3 )) −1)/(1−^3 (√2)+^3 (√4)))

$$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$

Question Number 202436    Answers: 2   Comments: 0

If α, β and γ are the roots of ax^3 + bx + c = 0 then frame an equation whose roots are α^2 , β^2 , γ^2 .

$$\mathrm{If}\:\alpha,\:\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{3}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{frame}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} ,\:\gamma^{\mathrm{2}} \:.\: \\ $$

Question Number 202419    Answers: 0   Comments: 0

Hard integral: Q202393

$$\mathrm{Hard}\:\mathrm{integral}:\:\mathrm{Q202393} \\ $$

Question Number 202418    Answers: 1   Comments: 0

Hard integral ∫∫∫∫∫∫∫∫∫ determinant ((a,b,c),(f,g,h),(j,k,l))dl dk dj dh dg df dc db da=

$$\mathrm{Hard}\:\mathrm{integral} \\ $$$$\int\int\int\int\int\int\int\int\int\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{f}}&{{g}}&{{h}}\\{{j}}&{{k}}&{{l}}\end{vmatrix}{dl}\:{dk}\:{dj}\:{dh}\:{dg}\:{df}\:{dc}\:{db}\:{da}= \\ $$

Question Number 202415    Answers: 2   Comments: 0

The value of ∫g′(x)f′(g(x))dx is...

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\int{g}'\left({x}\right){f}'\left({g}\left({x}\right)\right){dx}\:\mathrm{is}... \\ $$

Question Number 202408    Answers: 0   Comments: 0

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