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AlgebraQuestion and Answers: Page 46

Question Number 206063 Answers: 2 Comments: 0

If (x + (√(1 + x^2 )))(y + (√(1 + y^2 ))) = 1 then find (x + y)^2 .

$$\mathrm{If}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\left({y}\:+\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{find}\:\left({x}\:+\:{y}\right)^{\mathrm{2}} . \\ $$

Question Number 206025 Answers: 2 Comments: 0

2^(2024) = x (mod 10)

$$\:\:\:\:\:\mathrm{2}^{\mathrm{2024}} \:=\:{x}\:\left({mod}\:\mathrm{10}\right)\: \\ $$

Question Number 205990 Answers: 2 Comments: 0

If x = ((√3)/2) then (((√(1 + x)) + (√(1 − x)))/( (√(1 + x)) − (√(1 − x)))) = ?

$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{then}\:\frac{\sqrt{\mathrm{1}\:+\:{x}}\:+\:\sqrt{\mathrm{1}\:−\:{x}}}{\:\sqrt{\mathrm{1}\:+\:{x}}\:−\:\sqrt{\mathrm{1}\:−\:{x}}}\:=\:? \\ $$

Question Number 205988 Answers: 1 Comments: 0

Given f(x+1)=2^(f(x)) .f(1) and f(1)= 16 then f(2016)=?

$$\:\:{Given}\:{f}\left({x}+\mathrm{1}\right)=\mathrm{2}^{{f}\left({x}\right)} .{f}\left(\mathrm{1}\right) \\ $$$$\:\:{and}\:{f}\left(\mathrm{1}\right)=\:\mathrm{16}\: \\ $$$$\:\:{then}\:{f}\left(\mathrm{2016}\right)=? \\ $$

Question Number 205941 Answers: 1 Comments: 0

2+(2/(2−(2/(2+(2/(2−(2/(2+(2/3))))))))) =?

$$\:\:\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}−\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}}}}}\:=?\: \\ $$$$\: \\ $$

Question Number 205971 Answers: 2 Comments: 1

9 Mathematical Analysis ( I ) (X , d ) is a metric space and { p_n }_(n=1) ^∞ is a sequence in X such that , p_n →^(convergent) p . If , K= {p_n }_(n=1) ^∞ ∪ { p } then prove K , is compact in X .

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{9}\:\:\mathscr{M}{athematical}\:\:\:\:\mathscr{A}{nalysis}\:\left(\:{I}\:\right) \\ $$$$\:\:\left({X}\:,\:{d}\:\right)\:{is}\:{a}\:{metric}\:{space}\:{and}\:\:\:\:\:\:\: \\ $$$$\:\:\:\left\{\:{p}_{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty} {is}\:{a}\:{sequence}\:{in}\:{X}\: \\ $$$$\:\:\:\:\:{such}\:{that}\:,\:{p}_{{n}} \overset{{convergent}} {\rightarrow}\:{p}\:.\:{If}\:,\:{K}=\:\left\{{p}_{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty} \cup\:\left\{\:{p}\:\right\}\:{then} \\ $$$$\:\:\:\:\:{prove}\:\:{K}\:,\:{is}\:{compact}\:{in}\:{X}\:.\: \\ $$$$\:\:\:\: \\ $$

Question Number 205914 Answers: 4 Comments: 2

if a+b+c+d+e+f=10 and a^2 +b^2 +c^2 +d^2 +e^2 +f^2 =25, find a_(min) and f_(max) .

$${if}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}\:{and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{find} \\ $$$${a}_{{min}} \:{and}\:{f}_{{max}} . \\ $$

Question Number 205885 Answers: 2 Comments: 0

Find: lim_(n→∞) ()^(1/n) (((2n)),(( n)) ) = ?

$$\mathrm{Find}:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{}\begin{pmatrix}{\mathrm{2n}}\\{\:\:\mathrm{n}}\end{pmatrix}\:=\:? \\ $$

Question Number 205884 Answers: 2 Comments: 1

Question Number 205817 Answers: 2 Comments: 0

2^( x + log_2 3) = 12 ⇒ find: x = ?

$$\mathrm{2}^{\:\boldsymbol{\mathrm{x}}\:+\:\boldsymbol{\mathrm{log}}_{\mathrm{2}} \:\mathrm{3}} \:=\:\mathrm{12}\:\:\Rightarrow\:\:\mathrm{find}:\:\:\mathrm{x}\:=\:? \\ $$

Question Number 205790 Answers: 0 Comments: 0

Question Number 205789 Answers: 0 Comments: 0

Question Number 205772 Answers: 2 Comments: 0

Question Number 205770 Answers: 0 Comments: 0

If x,y,z>0 and xyz = 1 Prove that: ((((√2)x)^2 )/((1+xz)(1+xy))) + ((((√2)y)^2 )/((1+yz)(1+xy))) + ((((√2)z)^2 )/((1+xz)(1+yz))) ≥ (3/2)

$$\mathrm{If}\:\:\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{xyz}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\left(\sqrt{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{yz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{yz}\right)}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Question Number 205767 Answers: 1 Comments: 0

a,b,c ∈ℜ^+ a+b+c=1 a^2 /(1+b+c) + b^2 /(1+a+c) + c^2 /(1+a+b)≥k find k max. hint : inequality cauchy schwarz

$$ \\ $$$${a},{b},{c}\:\in\Re^{+} \:\: \\ $$$${a}+{b}+{c}=\mathrm{1} \\ $$$$\:\:\:{a}^{\mathrm{2}} /\left(\mathrm{1}+{b}+{c}\right)\:+\:{b}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{c}\right)\:\:+\:{c}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{b}\right)\geqslant{k} \\ $$$${find}\:\:\:{k}\:{max}. \\ $$$${hint}\::\:{inequality}\:{cauchy}\:{schwarz} \\ $$$$ \\ $$

Question Number 205746 Answers: 1 Comments: 0

Question Number 205733 Answers: 1 Comments: 0

If a,b,c∈R^+ and a+b+c=6 Prove that: ((a^2 −4)/(4a^2 −9a + 6)) + ((b^2 −4)/(4b^2 −9b + 6)) + ((c^2 −4)/(4c^2 −9c + 6)) ≤ 0

$$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{6} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4a}^{\mathrm{2}} −\mathrm{9a}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4b}^{\mathrm{2}} −\mathrm{9b}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4c}^{\mathrm{2}} −\mathrm{9c}\:+\:\mathrm{6}}\:\leqslant\:\mathrm{0} \\ $$

Question Number 205726 Answers: 1 Comments: 0

101 is chosen arbitrarily from the numbers 1,2,3,...,199,200. Prove that two of these selected numbers can be found such that one is divisible by the other.

$$ \\ $$101 is chosen arbitrarily from the numbers 1,2,3,...,199,200. Prove that two of these selected numbers can be found such that one is divisible by the other.

Question Number 205680 Answers: 1 Comments: 0

solve ⌊x ⌋ + ⌊ x^2 ⌋ = ⌊ x^3 ⌋

$$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\lfloor{x}\:\rfloor\:+\:\lfloor\:{x}^{\mathrm{2}} \rfloor\:=\:\lfloor\:{x}^{\mathrm{3}} \:\rfloor \\ $$$$ \\ $$

Question Number 205672 Answers: 1 Comments: 0