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Question Number 202395 Answers: 0 Comments: 0

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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 202388 Answers: 1 Comments: 0

P rove that: ∫ (dx/(b^4 +2ax^2 +c))=((tan^(−1) ((((√2)(√a)x)/( (√(c+b^4 ))))))/( (√2)(√a)(√(c+b^4 ))))+C if a∙(c+b^4 )>0

$$\:\:\boldsymbol{{P}}\:\boldsymbol{{rove}}\:\boldsymbol{{that}}:\:\:\:\:\int\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{b}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{c}}}=\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}\right)}{\:\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}+\boldsymbol{{C}} \\ $$$$\boldsymbol{{if}}\:\:\boldsymbol{{a}}\centerdot\left(\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} \right)>\mathrm{0} \\ $$$$ \\ $$

Question Number 202383 Answers: 2 Comments: 0

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

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

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Question Number 202371 Answers: 2 Comments: 0

If A ∈ M_(2×2) , det(A )≠ 0 , A^( 3) = A^2 +A ⇒ Find the values of det (2A −I )

$$ \\ $$$$\:\:\:{If}\:\:\:\:{A}\:\in\:{M}_{\mathrm{2}×\mathrm{2}} \:,\:{det}\left({A}\:\right)\neq\:\mathrm{0} \\ $$$$\:\:\:\:,\:\:{A}^{\:\mathrm{3}} \:=\:{A}^{\mathrm{2}} \:+{A}\:\Rightarrow\:{Find}\:{the}\: \\ $$$$\:\:\:\:{values}\:{of}\:\:\:{det}\:\left(\mathrm{2}{A}\:−{I}\:\right) \\ $$$$ \\ $$

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))

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

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If n ≥ 2 and U_n = (3 + (√5))^n + (3 − (√5))^n then prove that U_(n + 1) = 6U_n − 4U_(n − 1) .

$$\mathrm{If}\:{n}\:\geqslant\:\mathrm{2}\:\mathrm{and}\:\mathrm{U}_{{n}} \:=\:\left(\mathrm{3}\:+\:\sqrt{\mathrm{5}}\right)^{{n}} \:+\:\left(\mathrm{3}\:−\:\sqrt{\mathrm{5}}\right)^{{n}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{U}_{{n}\:+\:\mathrm{1}} \:=\:\mathrm{6U}_{{n}} \:−\:\mathrm{4U}_{{n}\:−\:\mathrm{1}} \:. \\ $$

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Question Number 202324 Answers: 2 Comments: 0

Find: (((1/2) + 1 + (3/2) + ... + 16)/((1/4) + (2/4) + (3/4) + ... + 8))

$$\mathrm{Find}:\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:...\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:...\:+\:\mathrm{8}} \\ $$

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Question Number 202315 Answers: 1 Comments: 0

If x : y : z = a : b : c then show that (((a + b + c)/(x + y + z)))^3 = ((abc)/(xyz)) .

$$\mathrm{If}\:{x}\::\:{y}\::\:{z}\:=\:{a}\::\:{b}\::\:{c}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\right)^{\mathrm{3}} \:=\:\frac{{abc}}{{xyz}}\:. \\ $$

Question Number 202308 Answers: 0 Comments: 1

Question Number 202307 Answers: 3 Comments: 0

Show that ((a^3 + b^3 )/(a^2 + b^2 )) > ((a^2 + b^2 )/(a + b))

$$\mathrm{Show}\:\mathrm{that}\:\frac{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:>\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }{{a}\:+\:{b}} \\ $$

Question Number 202306 Answers: 2 Comments: 0

Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,... obtained by subtracting integer squares from natural numbers.

$$ \\ $$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,... obtained by subtracting integer squares from natural numbers.

Question Number 202303 Answers: 3 Comments: 0

If a_1 = 1 and a_1 ∙ a_2 ∙ ... ∙ a_n = n^2 Find: a_2 + a_(13) = ?

$$\mathrm{If}\:\:\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:\:\:\mathrm{and}\:\:\:\mathrm{a}_{\mathrm{1}} \:\centerdot\:\mathrm{a}_{\mathrm{2}} \:\centerdot\:...\:\centerdot\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}_{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{13}} \:=\:? \\ $$

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