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AlgebraQuestion and Answers: Page 65

Question Number 191961 Answers: 0 Comments: 0

Question Number 191950 Answers: 1 Comments: 0

(√a) = 1 + (1/a) find: a^2 −a−(√a) = ?

$$\sqrt{\mathrm{a}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\:\:\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\sqrt{\mathrm{a}}\:=\:? \\ $$

Question Number 191946 Answers: 1 Comments: 0

((fof^(−1) (5)+fof^(−1) (15))/(fof^(−1) (5)))=?

$$\frac{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)+\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{15}\right)}{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)}=? \\ $$

Question Number 191947 Answers: 2 Comments: 0

prove that (√(a+b(√(a−b(√(a+b(√(a−b(√(...)))))))))) = (((√(4a−3b^2 ))+b)/2)

$${prove}\:{that} \\ $$$$\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{...}}}}}\:\:=\:\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }+{b}}{\mathrm{2}} \\ $$

Question Number 191867 Answers: 1 Comments: 0

Prove that if u=f(x^3 +y^3 ),where f is arbitry function then x^2 (∂u/∂y) = y^2 (∂u/∂x)

$${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$

Question Number 191839 Answers: 1 Comments: 0

2^a = 3^b = 36^c then prove that ab = 2c(a + b).

$$\mathrm{2}^{{a}} \:=\:\mathrm{3}^{{b}} \:=\:\mathrm{36}^{{c}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ab}\:=\:\mathrm{2}{c}\left({a}\:+\:{b}\right). \\ $$

Question Number 191833 Answers: 3 Comments: 0

Question Number 191830 Answers: 0 Comments: 0

Question Number 191798 Answers: 2 Comments: 2

Question Number 191790 Answers: 1 Comments: 7

prove that ((2x−4)/(2∙3∙4))+((3x−5)/(3∙4∙5))+((4x−6)/(4∙5∙6))+.....+((100x−102)/(100∙101∙102))=((103)/(102))

$${prove}\:{that} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{4}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}+\frac{\mathrm{3}{x}−\mathrm{5}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}+\frac{\mathrm{4}{x}−\mathrm{6}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}+.....+\frac{\mathrm{100}{x}−\mathrm{102}}{\mathrm{100}\centerdot\mathrm{101}\centerdot\mathrm{102}}=\frac{\mathrm{103}}{\mathrm{102}} \\ $$$$ \\ $$

Question Number 191775 Answers: 2 Comments: 0

Question Number 191735 Answers: 3 Comments: 0

Verify that ┐(p→q)→(p∧^┐ q) is tautology using laws of algebra

$${Verify}\:{that} \\ $$$$\:\urcorner\left({p}\rightarrow{q}\right)\rightarrow\left({p}\wedge^{\urcorner} {q}\right)\:{is}\:{tautology}\:{using}\:{laws}\:{of} \\ $$$${algebra} \\ $$

Question Number 191753 Answers: 1 Comments: 0

Question Number 191717 Answers: 0 Comments: 0

Question Number 191716 Answers: 0 Comments: 0

Question Number 191706 Answers: 2 Comments: 0

Question Number 191675 Answers: 2 Comments: 2

Solve for x : (x − (1/x))^(1/2) + (1 − (1/x))^(1/2) = x

$$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\left({x}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:{x} \\ $$

Question Number 191663 Answers: 4 Comments: 0

2^a +4^b +8^c =328 find a,b and c when(a,b,c)is natual number

$$\mathrm{2}^{{a}} +\mathrm{4}^{{b}} +\mathrm{8}^{{c}} =\mathrm{328} \\ $$$${find}\:{a},{b}\:{and}\:{c} \\ $$$${when}\left({a},{b},{c}\right){is}\:{natual}\:{number} \\ $$

Question Number 191632 Answers: 0 Comments: 1

∫_0 ^∞ (√(tan θ)) dθ

$$\int_{\mathrm{0}} ^{\infty} \sqrt{\mathrm{tan}\:\theta}\:{d}\theta \\ $$

Question Number 191631 Answers: 1 Comments: 0

∫_0 ^∞ x^(1/2) e^(−x^2 ) dx

$$\int_{\mathrm{0}} ^{\infty} {x}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 191624 Answers: 1 Comments: 0

Question Number 191623 Answers: 2 Comments: 0

Question Number 191615 Answers: 1 Comments: 0

a + b + c = 0. Prove that, (a/(a^2 − bc)) + (b/(b^2 − ca)) + (c/(c^2 − ab)) = 0.

$${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that}, \\ $$$$\frac{{a}}{{a}^{\mathrm{2}} \:−\:{bc}}\:+\:\frac{{b}}{{b}^{\mathrm{2}} \:−\:{ca}}\:+\:\frac{{c}}{{c}^{\mathrm{2}} \:−\:{ab}}\:=\:\mathrm{0}. \\ $$

Question Number 191614 Answers: 0 Comments: 1

((α^(100) +β^(100) )/(α^(100) −β^(100) )) = (((−w)^(100) +(−w^2 )^(100) )/((−w)^(100) −(−w^2 )^(100) )) = ((w^(100) +w^(200) )/(w^(100) −w^(200) )) = ((1+w^(100) )/(1−w^(100 ) )) = ((1+w)/(1−w)) = (2/(2w)) = (1/w) =

$$\frac{\alpha^{\mathrm{100}} +\beta^{\mathrm{100}} }{\alpha^{\mathrm{100}} −\beta^{\mathrm{100}} }\:=\: \\ $$$$\frac{\left(−{w}\right)^{\mathrm{100}} +\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} }{\left(−{w}\right)^{\mathrm{100}} −\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} } \\ $$$$=\:\frac{{w}^{\mathrm{100}} +{w}^{\mathrm{200}} }{{w}^{\mathrm{100}} −{w}^{\mathrm{200}} } \\ $$$$=\:\frac{\mathrm{1}+{w}^{\mathrm{100}} }{\mathrm{1}−{w}^{\mathrm{100}\:} } \\ $$$$=\:\frac{\mathrm{1}+{w}}{\mathrm{1}−{w}}\:=\:\frac{\mathrm{2}}{\mathrm{2}{w}}\:=\:\frac{\mathrm{1}}{{w}}\:=\: \\ $$

Question Number 191610 Answers: 1 Comments: 0

Q: if x+(1/x)=2cos(θ) prove it x^n +(1/x^n )=2cos(nθ)

$${Q}:\:{if}\:\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{2}{cos}\left(\theta\right)\:\:{prove}\:{it}\:\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=\mathrm{2}{cos}\left({n}\theta\right) \\ $$

Question Number 191589 Answers: 1 Comments: 1

a^x = bc, b^y = ca, c^z = ab. Prove that, (x/(1 + x)) + (y/(1 + y)) + (z/(1 + z)) = 2. (Without using log) a ≠ b ≠ c

$${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$$\left(\mathrm{Without}\:\mathrm{using}\:\mathrm{log}\right) \\ $$$${a}\:\neq\:{b}\:\neq\:{c} \\ $$

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