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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm GTNN-LN 2. B = 4x² – 4x – 3 3. C = x² – x 4. D = -9x² + 5x – 1 5. E = 2x² – 2xy + y² +6x +2029 6. Cho a + b = 1 Hãy Tìm: MinS = a²

Toán Lớp 8: Tìm GTNN-LN
2. B = 4x² – 4x – 3
3. C = x² – x
4. D = -9x² + 5x – 1
5. E = 2x² – 2xy + y² +6x +2029
6. Cho a + b = 1 Hãy Tìm:
MinS = a² + b²
MaxP = ab

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    $2.B=4x^2-4x-3$
    $=4x^2-4x+1-4$
    $=(2x-1)^2-4\geq-4$
    Dấu $”=”$ xảy ra khi $x=\frac12$
    $MIN_B=-4\ \text{khi}\ x=\frac12$
    $3.C=x^2-x$
    $=x^2-x+\frac14-\frac14$
    $=(x-\frac12)^2-\frac14\geq-\frac14$
    Dấu $”=”$ xảy ra khi $x=\frac12$
    $MIN_C=-\frac14\ \text{khi}\ x=\frac12$
    $4.D=-9x^2+5x-1$
    $=-(9x^2-5x+\frac{25}9)+\frac{16}9$
    $=-(3x-\frac53)^2+\frac{16}9\leq\frac{16}9$
    Dấu $”=”$ xảy ra khi $x=\frac59$
    $MAX_B=\frac{16}9\ \text{khi}\ x=\frac59$
    $5.E=2x^2-2xy+y^2+6x+2029$
    $=x^2+6x+9+x^2-2xy+y^2+2020$
    $=(x+3)^2+(x-y)^2+2020\geq2020$
    Dấu $”=”$ xảy ra khi $x=2020$
    $MIN_B=2020\ \text{khi}\ x=-3;y=-x=3$
    Mik ko làm được bài 6, sorry bạn
    Xin hay nhất!!!

  2. Giải đáp:
    2)B=4x^2-4x-3
    =4x^2-4x+1-4
    =(2x-1)^2-4>=-4
    Dấu “=” xảy ra khi x=1/2
    3)C=x^2-x
    =x^2-x+1/4-1/4
    =(x-1/2)^2-1/4>=-1/4
    Dấu “=” xảy ra khi x=1/2
    4)D=-9x^2+5x-1
    =-9(x^2-5/9x)-1
    =-9(x^2-5/9x+25/324-25/324)-1
    =-9(x-5/18)^2+25/36-1
    =-9(x-5/18)^2-11/36<=-11/36
    Dấu “=” xảy ra khi x=5/18
    E=2x^2-2xy+y^2+6x+2029
    =x^2-2xy+y^2+x^2+6x+9+2020
    =(x-y)^2+(x+3)^2+2020>=2020
    Dấu “=” xảy ra khi x=y=-3
    6)a+b=1
    S=a^2+b^2
    <=>2S=2a^2+2b^2
    <=>2S=a^2-2ab+b^2+a^2+2ab+b^2
    <=>2S=(a-b)^2+(a+b)^2
    <=>2S=(a-b)^2+1>=1
    =>S>=1/2
    Dấu “=” xảy ra khi a=b=1/2
    P=ab
    <=>4P=4ab
    <=>4P=a^2+2ab+b^2-(a^2-2ab+b^2)
    <=>4P=(a+b)^2-(a-b)^2
    <=>4P=1-(a-b)^2<=1
    =>P<=1/4
    Dấu “=” xảy ra khi a=b=1/2

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222-9+11+12:2*14+14 = ? ( )