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Answers ( 459 )
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    a - a always equal 0

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    The correct answer is C

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    We have: 

    \(a^2+b^2+c^2\ge ab+bc+ca\)

    The "=" happen when and only when a = b = c

    Applying the Cauchy-Schwarz inequality, we have:

    \(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\le\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2+2+2}\le\dfrac{9}{ab+bc+ca+6}=\dfrac{9}{9}=1\)

    So......................

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    \(\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{\dfrac{17}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{4}{17}}}=\dfrac{1}{4+\dfrac{1}{\dfrac{72}{17}}}=\dfrac{1}{4+\dfrac{17}{72}}=\dfrac{1}{\dfrac{305}{72}}=\dfrac{72}{305}\)

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    \(\left(\left(X^{\dfrac{1}{2}}\right)^{\dfrac{1}{2}}\right)^{\dfrac{1}{2}}=\sqrt{\sqrt{\sqrt{X}}}=\sqrt[2^3]{X}=\sqrt[8]{X}\)

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    The fraction in odd location have the denominator is sequence of odd numbers from 3

    The fraction in even location have the denominator is sequence of 2n from 21

    And their numerator is 1

    So the fraction in ? is \(\dfrac{1}{13}\)

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    Their average is: \(\dfrac{6+7+42}{16}=\dfrac{55}{16}=3,4375\)

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    \(\dfrac{30\%\cdot\dfrac{25}{17}}{\dfrac{3}{4}\cdot\dfrac{1}{2}}=\dfrac{\dfrac{15}{34}}{\dfrac{3}{8}}=\dfrac{20}{17}\)

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    200 days = 28 weeks and 4 days

    So 200 days from now will be 3 + 4 = 7 or Saturday 

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    Call a zeeko is x, a teeko is y. We have:

    \(\dfrac{1}{2}x=60\%y\)

    \(\Leftrightarrow0,5x=0,6y\)

    \(\Leftrightarrow x=1,2y\)

    \(\Leftrightarrow\dfrac{x}{y}=1,2\)

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    With this post, we have:

    130%A = B

    60%B = C

    120%C = D

    => D = 60% of 120% of B

    => D = 72%B

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    \(\dfrac{1}{10\sqrt{10}}=\dfrac{1}{\sqrt{10^3}}\)

    \(\dfrac{\sqrt{10}}{\sqrt{10^4}}=\dfrac{1}{\sqrt{10^3}}\)

    So A = B

    The answer is c

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    \((x-1)(2x-1) = 9-x\)

    => \(2x^2 - 3x + 1 = 9- x\)

    => \(2x^2 - 2x = 8\)

    => \(2x^2 - 2x - 8 = 0\)

    => \(2(x^2 - x - 4)= 0\)

    => \(2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{17}{2}=0\)

    => \(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{2}\)

    => \(\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{4}\)

    => \(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{1}{2}=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

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    3rd term is: 1 + 2 = 3

    4th term is: 1 + 2 + 3 = 6

    5th term is: 1 + 2 + 3 + 6 = 12

    ....

    15th term is: 1 + 2 + 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = 1536 

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    \(0.2\overline{36}=\dfrac{236-2}{990}=\dfrac{234}{990}=\dfrac{13}{55}\)

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    Mr. Bee 

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    People call this triangle is Pascal.

                                                      1

                                            1                    1

                                  1                    2                    1

                        1                    3                    3                    1

              1                    4                    6                    4                    1

    1                    5                    10                    10                    5                    1

    So the answer is 1,5,10,10,5,1

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    This sequence is ...,5k,6k,7k,8k,9k,10k,11k,12k,... with k = 4

    So each terms in this will \(⋮4\)

    So in 4 answers, answer d is correct :]

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    This sequence is 0k,1k,2k,3k,4k,5k,6k,7k,8k,...,300k with k = 3

    So the missing numbers in "..." is 9k to 299k

    So there are: 299 - 9 + 1 = 291 numbers in this

  • See question detail

    O 3 2 17 11 13 A B C

    \(S_{ABC}=\dfrac{\left(13-11\right)\left(17-2\right)}{2}=\dfrac{2\cdot15}{2}=15\left(unit^2\right)\)

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Questions ( 41 )
  • Given: a + b + 2c = 1.

    Find the minimum and maximum of A = ab - 4bc - 4ca

  • 5. Given the triangle ABC balance at A (AB = AC). On BC take M any such that BM < CM. From M draw parallel lines with AC cut AB at E and parallel with AB cut AC at F. Let N be the symmetry point of M over EF.
    a) Calculate the perimeter of AEMF. Know: AB = 7cm
    b) Prove that: AFEN is a trapezoid trait
    c) Calculated: \(\widehat{ANB}+\widehat{ACB}\)
    d) Where M is the AEMF quadrilateral is diamond and needs to be added to the triangle ABC so that the AEMF is square.

  • 4. (easy) Via the center of the triangle G, the straight line parallel to AC, cut AB and BC respectively at M and N. Calculate the length of the field, know AM + NC = 16 (cm); Triangle triangle ABC equals 75 cm

  • 3. Given triangle ABC. On BC, CA, AB take D, E, F so that AD, BE, DF intersect at H.

    Prove that:

    a, \(\dfrac{AH}{AD}+\dfrac{BH}{BE}+\dfrac{CH}{CF}=2\)

    b, \(\dfrac{AH}{HD}+\dfrac{BH}{HE}+\dfrac{CH}{HF}\ge6\)

     

  • 2. Given triangle ABC. M is a point in the interior of triangle ABC. Let D, E, F be the midpoints AB, AC, BC; A ', B', C 'is the symmetry point of M through F, E, D.
    a, CMR: AB'A'B is a parallelogram.
    b, CMR: CC 'goes through the midpoint of AA'

  • In this web - Mathulike mostly is algebraic and only few geometry. So I will post some exercises of geometry.

    1. Given ABCD rectangle with AB = 2AD, call E, I in turn is the midpoint of AB and CD. Connect D to E. Draw the Dx beam perpendicular to DE, the Dx beam to the opposite ray of the CB ray at M. On the opposite side of the CE beam take the K point so that DM = EK. Let G be the intersection of DK and EM.
    a) Calculate DBK angle.
    b) Let F be the right-angled foot from K to BM. Prove that four points A, I, G, H are in the same line.

    undefined

  • In recent times, I have seen some questions not related to mathematics. And that is not good. I suspect there are some accounts where the main account is https://e-learning.codienhanoi.edu.vn/profile/ducanh2007 has posted the question itself and tick yourself. As an coodinator, I hope Admin will soon resolve this issue. And I advise you to have the nickname "Cristiano Ronaldo" stop ticking yourself. I hope Mr.Bee will solve it satisfactorily.

    And to stop this status, I will post some question in my free time. haha

  • With all a,b.

    Prove that: a, \(\left(a+b\right)^2\ge4ab\)

                       b, \(a+b\ge2\sqrt{ab}\)

    Simple haha

  • With all a,b

    Prove that: \(a^4+b^4\ge a^3b+b^3a\)

  • Given: \(a,b\in Z\)

    Prove that: \(\left(a+b+c\right)^3\ge3\left(ab+bc+ca\right)\)

     

  • Given: a,b > 0

    Prove that: \(\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\)

  • Given: x + y = 2 

    Prove that: \(a^4+b^4\ge2\)

  • Given: a + b > 1

    Prove that: \(a^4+b^4>\dfrac{1}{8}\)

  • A B C D E F

    Prove that:

    \(\dfrac{3}{4}\left(AB+AC+BC\right)< AD+BE+CF< AB+AC+BC\)

     

  • Caculator: 

    \(A=\dfrac{\dfrac{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}}}{\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}}{\dfrac{1}{2016}+\dfrac{2}{2015}+...+\dfrac{2016}{1}}}\)

  • Given: (2x1 - 5x1)2018 + (2x2 - 5x2)2018 + ... + (2x2019 - 5x2019)2018 \(\le\)0

    Prove that: \(\dfrac{x_1+x_2+...+x_{2019}}{y_1+y_2+...+y_{2019}}=2,5\)

  • Find the minimum of P = |7x - 5y| + |2z - 3x| + |xy + yz + zx - 2000| 

  • Calculator:

    \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+2006}\right)\)

  • Question 2: Prove that if a,b,c and \(\sqrt{a}+\sqrt{b}+\sqrt{c}\) are rational numbers then \(\sqrt{a},\sqrt{b},\sqrt{c}\) are rational numbers

  • Today and next days, I will post some questions everyday. 

    Question 1: Find the integer part of \(\sqrt{2}+\sqrt[3]{\dfrac{3}{2}}+\sqrt[4]{\dfrac{4}{3}}+...+\sqrt[n+1]{\dfrac{n+1}{n}}\)

     

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