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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

Question Number 202406 Answers: 2 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 202395 Answers: 0 Comments: 0

Question Number 202393 Answers: 2 Comments: 0

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))).

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

Question Number 202376 Answers: 0 Comments: 1

Question Number 202374 Answers: 1 Comments: 2

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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