Question

Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $

score 105 · Answer 1 · 2016-09-02 18:20:29

105

Views

Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $

Published on 02 Sep 2016 - 18:20

score 284 · Answer 2 · 2016-09-03 21:01:15

284

Views

limit of product of $(a_1a_2.\dots a_n)^{\frac{1}{n}}$

Published on 03 Sep 2016 - 21:01

score 40 · Answer 3 · 2016-09-05 01:15:44

40

Views

Show that the number of solutions is more than $1$

Published on 05 Sep 2016 - 1:15

score 825 · Answer 4 · 2016-09-05 17:27:11

825

Views

Limits; factoring: $\lim_{x \to 2} \frac {\sqrt {x+2}-2}{x-2}$

Published on 05 Sep 2016 - 17:27

score 485 · Answer 5 · 2016-09-05 18:21:36

485

Views

Relation between signs of imaginary parts of $(x+iy)$ and $\sqrt{x+iy}$

Published on 05 Sep 2016 - 18:21

score 374 · Answer 6 · 2026-03-29 15:21:29

374

Views

Digit after decimal point of radicals

Published on 29 Mar 2026 - 15:21

score 125 · Answer 7 · 2016-09-07 01:40:04

125

Views

Limits with square roots

Published on 07 Sep 2016 - 1:40

score 4.7k · Answer 8 · 2026-03-30 01:29:45

4.7k

Views

How do I express 14/(3-√2) in the form of b+c√d?

Published on 30 Mar 2026 - 1:29

score 667 · Answer 9 · 2026-03-25 03:03:08

667

Views

How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$

Published on 25 Mar 2026 - 3:03

score 86 · Answer 10 · 2016-09-09 11:09:02

86

Views

Square roots: $ -\sqrt{a} = \sqrt{b}$

Published on 09 Sep 2016 - 11:09

score 114 · Answer 11 · 2026-03-25 09:29:05

114

Views

${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$

Published on 25 Mar 2026 - 9:29

score 178 · Answer 12 · 2016-09-11 00:10:16

178

Views

Why don't $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds true when one of $a$ & $b$ is imaginary number?

Published on 11 Sep 2016 - 0:10

score 163 · Answer 13 · 2026-03-26 01:16:03

163

Views

Is $(5+\sqrt[3]2)^n$ ever an integer for $n \in \Bbb Z \setminus \{0\}$?

Published on 26 Mar 2026 - 1:16

score 74 · Answer 14 · 2016-09-12 13:23:19

74

Views

What does an idiosyncratic solution in radicals means?

Published on 12 Sep 2016 - 13:23

score 123 · Answer 15 · 2016-09-12 17:16:05

123

Views

Find $\lim_{n \to \infty} \sqrt[n]{1+2^{(-1)^n}}$ and $\lim_{n \to \infty} \sqrt[n]{1^k+2^k+\cdots+n^k}$

Published on 12 Sep 2016 - 17:16

List Question

Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $

limit of product of $(a_1a_2.\dots a_n)^{\frac{1}{n}}$

Show that the number of solutions is more than $1$

Limits; factoring: $\lim_{x \to 2} \frac {\sqrt {x+2}-2}{x-2}$

Relation between signs of imaginary parts of $(x+iy)$ and $\sqrt{x+iy}$

Digit after decimal point of radicals

Limits with square roots

How do I express 14/(3-√2) in the form of b+c√d?

How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$

Square roots: $ -\sqrt{a} = \sqrt{b}$

${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$

Why don't $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds true when one of $a$ & $b$ is imaginary number?

Is $(5+\sqrt[3]2)^n$ ever an integer for $n \in \Bbb Z \setminus \{0\}$?

What does an idiosyncratic solution in radicals means?

Find $\lim_{n \to \infty} \sqrt[n]{1+2^{(-1)^n}}$ and $\lim_{n \to \infty} \sqrt[n]{1^k+2^k+\cdots+n^k}$

Trending Questions

Popular # Hahtags

Popular Questions