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Answers ( 5 )
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    Sơ đồ dưới đây cho thấy năm hình vuông kích cỡ bằng nhau. phần bóng mờ của mỗi hình bị loại bỏ.Các con số kết quả, có chu vi là tương đương với chu vi của hình vuông bị cắt

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    No apples, because PEAR doesn't grow out of apples.

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    a, consider the triangle AHFK: \(\widehat{A}=\widehat{H}=\widehat{K}=90^o\)

    should AHFK is rectangle

    b,  consider the triangle ACF :OA=OC;EC=EF

    should OE is the average line of the triangle ACF

    should OE//AF or AF//BD

    similar: the average line is EJ triangle ACF 
    should EJ//AC

    similar: the average line is EJ triangle ACF 
    should EJ//AC

    \(\Rightarrow\widehat{AKJ}=\widehat{KAJ}+\widehat{KAJ}=\widehat{KDE}\)pairs of isotopes

    \(\Rightarrow\widehat{AKJ}=\widehat{KDE}\) or KDE loss triangle

    inferred :\(\widehat{JK}=\widehat{DEK}=\dfrac{180-KDE}{2}\) should K; J and E line 
    that K; J; H line 
    should K; H and E also in line and HK//AC

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    A B D C K F J H E O

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    x+y=1             \(\Rightarrow\)\(x^2+2xy+y^2=1\)\(\left(1\right)\)

    which \(\left(x-y\right)^2\ge0\)     else \(x^2-2xy+y^2\ge0\left(2\right)\)

    plus 1 and 2 we have   \(2\left(x^2+y^2\right)\ge1\Rightarrow x^2+y^2\ge\dfrac{1}{2}\)

    minA=\(\dfrac{1}{2}\)      When and only when  x = y = \(\dfrac{1}{2}\)

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Questions ( 13 )
  • Search GTLN
    a) 4 x-x ^ 2-5
    b) 5-x ^ 2-6 x

     

     

  • Proof: a) x ^ 2 + x + 1 > 0 for all x; 
    b) 2 x ^ 2 + 2 x + 1 > = 0 for all x

  • Search GTNN
    a) x ^ 2-4 x + 1
    b) 3 x ^ 2-6 x-1
    c) 4 x ^ 2 + 4 x-2
    Search GTLN
    a) 4 x-x ^ 2-5
    b) 5-x ^ 2-6 x

  •  1

     

    for triangle ABC in A balance. guys that BH is perpendicular to AC. D is the midpoint of BC. On the beam of rays DH took the point M to DM = DH. Proven 
    a is, triangle BMD = triangle CHD
    b, BC is the ray of ABM corner stool
    c, suppose x BH HC.so > comparison of two go cs BHD and CHD

  • for triangle ABC in A balance. guys that BH is perpendicular to AC. D is the midpoint of BC. On the beam of rays DH took the point M to DM = DH. Proven 
    a is, triangle BMD = triangle CHD
    b, BC is the ray of ABM corner stool
    c, suppose x BH HC.so > comparison of two go cs BHD and CHD

  • For the triangle loss in A = 90 degrees

    M point in BC. drawing-MO perpendicular to AB

    MK drawing perpendicular to ac. On the beam of rays of OM and KM in turn retrieved 2 points D and E for OM = OD, KM = KE

    C/M AD = AE

  • phân tích đa thức thành nhâ tử

    1,x^3−x^2+x−1

    2,6x^2y−2xy^2+3x−y

    3,4x^2+1

    4,x^2−9x+8

    5,x^3−2x^2y+3xy^2

    6,x^2−6x+y−y^2

    7,x^2−xy−2x+2y

  • for 3\(a^2+3b^2=10ab\)

    because b>a >0

    the provincial P=\(\dfrac{a-b}{a+b}\)
     
  • Search x know
    1,  \(x^3-25x=0\)
    2,  \(2x\left(x-1\right)-3x+3=0\)
    3,  \(x^2-36+\left(x^2-12x+36\right)=0\)
  • analysis of the factor

    1,\(x^3+2x^2+x-2\)

    2,\(x^2-2x+5\)

    3,\(a^3+4a^2+4a+3\)

    4,\(4a^2b^2-\left(a^2-b^2\right)^2\)

    5,\(a^3+6a^2+8+12a\)

    6,\(4+4x-x^2\)

    7,\(\left(2x-5\right)^2-\left(5x-3x\right)\)

     

  • Search x know

    1,\(x\left(x-3\right)+x-3=0\)

    2,\(5x\left(x-2\right)-x+2=0\)

  • 1,analysis of the factor

    zx-zy-5x+5y

    \(x^2-y+x^2y-x^3\)

    \(4x^2-t^2-4ty-4y^2\)

    \(4x-x^2-4+y^2\)

    \(y^2-x^2-2x-1\)

     

  • 1, a school has 550 female students. Eagle boys 10/11 of the female student. ask the school which have all how many students?

    2, men have 42 volumes. for the average 2/7 of breaks, to left after rupture of 3/5 for peace. ask the average and how much the washhouse you get each term booklet

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