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Answers ( 5 )
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    bị ẩn mất rồi nói chung là bn ấy kêu gọi \(\text{}\)

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    bạn ấy còn nhắn với tôi là "hi, bạn cho mk nha"

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    Kaya Renger You're a cheater too. Do you know the word "Tự hỏi tự trả lời" in Vietnam ?

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    AP Statistics: The Best Test Preparation for the Advanced Placement Exam - Jack J. Berger, Research and Education Association - Google Sách

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     positive integers less than 10 are: {1,2,3,4,5,6,7,8,9} 
    So if you pick 2 of them, you end up with a subset of integers from 12 through 98, and there are 9 x 8 = 72 possible combinations (no 10's and no multiples of 11). 

    The perfect squares between 12 and 98 are 16, 25, 36, 49, 64, and 81. 

    So there are 6 out of 72 possibilities, 6/72 = 1/12 = 8.3%

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Questions ( 3 )
  • For \(\Delta ABC\) right-angled triangle at  \(A\) have \(AB=AC\), uppose in the triangle that the point M satisfies \(\widehat{MBA}=\widehat{MAC}=\widehat{MCB}\). Calculate the ratio of \(MA: MB: MC\)

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  • Given \(a,b,c\) are non-negative numbers such that \(ab+bc+ca=1\). Find the minimize value of \(P=\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\)

    -Source: Câu hỏi của michelle holder - Toán lớp 10 | Học trực tuyến (Ace Legona's solution is wrong)

  • Given $a,b,c>0$. Prove that $\frac{a^3b}{3a+b}+\frac{b^3c}{3b+c}+\frac{c^3a}{3c+a}\ge \frac{a^2bc}{2a+b+c}+\frac{ab^2c}{a+2b+c}+\frac{abc^2}{a+b+2c}$

    *)Source: https://olm.vn/hoi-dap/question/999979.html

    I tried AM-GM but unsuccess 

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