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AlgebraQuestion and Answers: Page 46

Question Number 202605 Answers: 1 Comments: 3

Question Number 202604 Answers: 1 Comments: 0

If x = (((√(a + 1)) + (√(a − 1)))/( (√(a + 1)) − (√(a − 1)))) and y = (((√(a + 1)) − (√(a − 1)))/( (√(a + 1)) + (√(a − 1)))) then show that ((x^2 − xy + y^2 )/(x^2 + xy + y^2 )) = ((4a^2 − 3)/(4a^2 − 1)) .

$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{and}\: \\ $$$${y}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} \:−\:{xy}\:+\:{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{1}}\:. \\ $$

Question Number 202551 Answers: 0 Comments: 1

Σ_(i=1) ^n (x+y)_i =(x+y)_1 +(x+y)_2 +...(x+y)_n =x_1 +y_1 +x_2 +y_2 +...+x_n +y_n please it′s correct ?

$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +...\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +...+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:? \\ $$$$ \\ $$

Question Number 202540 Answers: 1 Comments: 0

If (b/(a + b)) = ((3a − b − c)/(2b + c − a)) = ((3c − a)/(2a − b + 3c)) [a + b + c ≠ 0] then show that a = b = c.

$$\mathrm{If}\:\frac{{b}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{3}{a}\:−\:{b}\:−\:{c}}{\mathrm{2}{b}\:+\:{c}\:−\:{a}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{\mathrm{2}{a}\:−\:{b}\:+\:\mathrm{3}{c}}\: \\ $$$$\left[{a}\:+\:{b}\:+\:{c}\:\neq\:\mathrm{0}\right]\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:{a}\:=\:{b}\:=\:{c}. \\ $$

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 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 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 202464 Answers: 0 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 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 202408 Answers: 0 Comments: 0

Question Number 202400 Answers: 2 Comments: 1

Solve for a, b and c (1/a) + (1/(b + c)) = (1/2) (1/b) + (1/(c + a)) = (1/3) (1/c) + (1/(a + b)) = (1/4)

$$\mathrm{Solve}\:\mathrm{for}\:{a},\:{b}\:\mathrm{and}\:{c} \\ $$$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}\:+\:{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Question Number 202392 Answers: 2 Comments: 11

If the ratio of the roots of ax^2 + bx + b = 0 is p : q then show that (√(p/q)) + (√(q/p)) + (√(b/a)) = 0.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{b}\:=\:\mathrm{0} \\ $$$$\mathrm{is}\:{p}\::\:{q}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\sqrt{\frac{{p}}{{q}}}\:+\:\sqrt{\frac{{q}}{{p}}}\:+\:\sqrt{\frac{{b}}{{a}}}\:=\:\mathrm{0}. \\ $$

Question Number 202390 Answers: 0 Comments: 0

Question Number 202383 Answers: 2 Comments: 0

Show that ((a(√b) − b(√a))/(a(√b) + b(√a))) = (1/(a − b))(a + b − 2(√(ab))).

$$\mathrm{Show}\:\mathrm{that}\:\frac{{a}\sqrt{{b}}\:−\:{b}\sqrt{{a}}}{{a}\sqrt{{b}}\:+\:{b}\sqrt{{a}}}\:=\:\frac{\mathrm{1}}{{a}\:−\:{b}}\left({a}\:+\:{b}\:−\:\mathrm{2}\sqrt{{ab}}\right). \\ $$

Question Number 202359 Answers: 1 Comments: 0

2 , 8 , 32 , ... geometfic serie for b_m > 1024 find min(m) = ?

$$\mathrm{2}\:,\:\mathrm{8}\:,\:\mathrm{32}\:,\:...\:\mathrm{geometfic}\:\mathrm{serie} \\ $$$$\mathrm{for}\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{m}}} \:>\:\mathrm{1024}\:\:\:\mathrm{find}\:\:\:\mathrm{min}\left(\mathrm{m}\right)\:=\:? \\ $$

Question Number 202356 Answers: 4 Comments: 0

Find: 1−(sin30°)^2 + (sin30°)^4 − (sin30°)^6 + ...

$$\mathrm{Find}: \\ $$$$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}} \:+\:\left(\mathrm{sin30}°\right)^{\mathrm{4}} \:−\:\left(\mathrm{sin30}°\right)^{\mathrm{6}} \:+\:... \\ $$

Question Number 202353 Answers: 2 Comments: 0

If 2x = a + (√((4b^3 − a^3 )/(3a))) and 2y = a − (√((4b^3 − a^3 )/(3a))) then show that x^3 + y^3 = b^3 .

$$\mathrm{If}\:\mathrm{2}{x}\:=\:{a}\:+\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}}\:\mathrm{and} \\ $$$$\mathrm{2}{y}\:=\:{a}\:−\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:=\:{b}^{\mathrm{3}} \:. \\ $$

Question Number 202352 Answers: 0 Comments: 9

Question Number 202348 Answers: 1 Comments: 0

Let a,b,c ∈R^+ , a+b+c=3 prove the following inequality (((2a−3)^2 )/b)+(((2b−3)^2 )/c)+(((2c−3)^2 )/a)≥((a^2 +b^2 )/(a+b))+((b^2 +c^2 )/(b+c))+((c^2 +a^2 )/(c+a))

$$ \\ $$$$\mathrm{Let}\:{a},{b},{c}\:\:\in\mathbb{R}^{+} \:,\:{a}+{b}+{c}=\mathrm{3}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality} \\ $$$$\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}+\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} }{{c}}+\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)^{\mathrm{2}} }{{a}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}+{a}} \\ $$

Question Number 202340 Answers: 2 Comments: 0

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