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Question Number 61510 Answers: 0 Comments: 0

S_1 =Σ_(k=1) ^n (√((16n−16k)(16n+16k))) S_2 =Σ_(k=1) ^n (√((16k−16)(16k+16))) lim_(n→∞) ((S_1 +S_2 )/n^2 )=?

$${S}_{\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{n}−\mathrm{16}{k}\right)\left(\mathrm{16}{n}+\mathrm{16}{k}\right)} \\ $$$${S}_{\mathrm{2}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{k}−\mathrm{16}\right)\left(\mathrm{16}{k}+\mathrm{16}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} }{{n}^{\mathrm{2}} }=? \\ $$

Question Number 61490 Answers: 0 Comments: 10

(√(a−(√(a+x))))+(√(a+(√(a−x))))=2x this is the solution Sir Aifour and me found trivial solution a=x=0 a, x ∈R x=((√2)/8)(r+(√(r^2 +4)))(√(2(4a−1)−r^2 −r(√(r^2 +4)))) with r=−2(√((4a−3)/3))sin ((1/3)arcsin ((3(√3))/((4a−3)^(3/2) ))) no solution for a<a_0 with a_0 ≈1.509830340886

$$\sqrt{{a}−\sqrt{{a}+{x}}}+\sqrt{{a}+\sqrt{{a}−{x}}}=\mathrm{2}{x} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{Sir}\:\mathrm{Aifour}\:\mathrm{and}\:\mathrm{me}\:\mathrm{found} \\ $$$$ \\ $$$$\mathrm{trivial}\:\mathrm{solution}\:{a}={x}=\mathrm{0} \\ $$$$ \\ $$$${a},\:{x}\:\in\mathbb{R} \\ $$$$ \\ $$$${x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left({r}+\sqrt{{r}^{\mathrm{2}} +\mathrm{4}}\right)\sqrt{\mathrm{2}\left(\mathrm{4}{a}−\mathrm{1}\right)−{r}^{\mathrm{2}} −{r}\sqrt{{r}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\mathrm{with} \\ $$$${r}=−\mathrm{2}\sqrt{\frac{\mathrm{4}{a}−\mathrm{3}}{\mathrm{3}}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\left(\mathrm{4}{a}−\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right) \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a}<{a}_{\mathrm{0}} \:\mathrm{with}\:{a}_{\mathrm{0}} \approx\mathrm{1}.\mathrm{509830340886} \\ $$

Question Number 61565 Answers: 2 Comments: 3

x(y+1) + y(x−1) = 12 x(√y) + y(√x) = 21 x, y ∈ R x + y = ?

$${x}\left({y}+\mathrm{1}\right)\:+\:{y}\left({x}−\mathrm{1}\right)\:\:=\:\:\mathrm{12} \\ $$$${x}\sqrt{{y}}\:\:+\:\:{y}\sqrt{{x}}\:\:\:\:=\:\:\mathrm{21} \\ $$$${x},\:{y}\:\:\in\:\:\mathbb{R} \\ $$$${x}\:+\:{y}\:\:=\:\:? \\ $$

Question Number 61479 Answers: 3 Comments: 1

Solve for x: ((6(√(2x)))/(x − 1)) + ((5(√(x − 1)))/(2x)) = 13

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\mathrm{6}\sqrt{\mathrm{2x}}}{\mathrm{x}\:−\:\mathrm{1}}\:+\:\frac{\mathrm{5}\sqrt{\mathrm{x}\:−\:\mathrm{1}}}{\mathrm{2x}}\:\:\:=\:\:\mathrm{13} \\ $$

Question Number 61470 Answers: 1 Comments: 0

Is there any other solution besides {x=a,y=b} or {x=b,y=a} of the following system of equations x+y=a+b ∧ x^7 +y^7 =a^7 +b^7 ?

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{besides} \\ $$$$\left\{\mathrm{x}=\mathrm{a},\mathrm{y}=\mathrm{b}\right\}\:\mathrm{or}\:\left\{\mathrm{x}=\mathrm{b},\mathrm{y}=\mathrm{a}\right\}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}\:\:\wedge\:\mathrm{x}^{\mathrm{7}} +\mathrm{y}^{\mathrm{7}} =\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} \:\:? \\ $$$$ \\ $$

Question Number 61465 Answers: 1 Comments: 1

∫_0 ^(2π) (1/(a^2 cos^2 (t)+b^2 sin^2 (t)))dt=((2π)/(ab))?