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

Toán Lớp 8: 1.Tìm Min A=2x^2-6x-1 B=x^2-x-2 2.Tìm Max A=3-x^2-2x B=2x-3x^2-4

Tháng Mười Một 15, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: 1.Tìm Min
A=2x^2-6x-1
B=x^2-x-2
2.Tìm Max
A=3-x^2-2x
B=2x-3x^2-4

Comments ( 2 )

  1. ngoclanquynh
    ngoclanquynh
    15 Tháng Mười Một, 2022 at 10:31 chiều
    Reply
    Câu 1:
    a)
    A = 2x^2 – 6x – 1
    =  2 (x^2 – 3x + 9/4) – 11/2
    = 2 (x – 3/2)^2 – 11/2
    \forall x ta có :
    (x-3/2)^2 \ge 0
    => 2 (x-3/2)^2 \ge 0
    => 2 (x-3/2)^2 – 11/2 \ge -11/2
    => A \ge -11/2
    Dấu = xảy ra <=> x – 3/2 = 0
    <=> x = 3/2
    Vậy \text{Min}_A = -11/2 <=> x = 3/2
    b)
    B = x^2 – x – 2
    = (x^2 – x+  1/4) – 9/4
    =  (x – 1/2)^2 – 9/4
    \forall x ta có :
    (x-1/2)^2 \ge 0
    => (x-1/2)^2 – 9/4 \ge -9/4
    => B \ge -9/4
    Dấu = xảy ra <=>x-1/2=0
    <=>x=1/2
    Vậy text{Min}_B = -9/4 <=> x = 1/2
    Câu 2:
    a)
    A = 3 – x^2 – 2x
    = – (x^2 + 2x + 1) + 4
    =  – (x+1)^2 + 4
    \forall x ta có :
    (x+1)^2 \ge 0
    => – (x+1)^2 \le 0
    => – (x+1)^2 + 4 \le 4
    => A \le 4
    Dấu = xảy ra <=>x+1=0
    <=> x =-1
    Vậy \text{Max}_A = 4 <=> x = -1
    b)
    B= 2x – 3x^2 – 4
    =  -3 (x^2 – 2/3x + 1/9) – 11/3
    = -3 ( x – 1/3)^2 -11/3
    \forall x ta có :
    (x-1/3)^2 \ge 0
    => -3 (x-1/3)^2 \le 0
    => -3 (x-1/3)^2 – 11/3 \le -11/3
    => B \le -11/3
    Dấu = xảy ra <=> x- 1/3=0
    <=>x=1/3
    Vậy \text{Max}_B = -11/3 <=> x = 1/3

  2. ngochuong2k2
    ngochuong2k2
    15 Tháng Mười Một, 2022 at 10:31 chiều
    Reply
    1.
    A=2x²-6x-1
    =2(x²-3x-1/2)
    =2(x²-3x+9/4-11/4)
    =2(x²-3x+9/4)-11/2
    =2[x²-2.x. 3/2+(3/2)^2]-11/2
    =2(x-3/2)^2-11/2
    Ta có:(x-3/2)^2≥0∀x
    ⇒2(x-3/2)^2≥0∀x
    ⇒2(x-3/2)^2-11/2≥-11/2∀x
    Vậy A_(min)=-11/2 khi x-3/2=0⇔x=3/2
    B=x²-x-2
    =x²-x+1/4-9/4
    =(x²-x+1/4)-9/4
    =[x²-2.x. 1/2+(1/2)^2]-9/4
    =(x-1/2)^2-9/4
    Ta có:(x-1/2)^2≥0∀x
    ⇒(x-1/2)^2-9/4≥-9/4∀x
    Vậy B_(min)=-9/4 khi x-1/2=0⇔x=1/2
    2.
    A=3-x²-2x
    =-(x²+2x-3)
    =-(x²+2x+1-4)
    =-(x²+2x+1)+4
    =-(x+1)²+4
    Ta có:(x+1)²≥0∀x
    ⇒-(x+1)²≤0∀x
    ⇒-(x+1)²+4≤4∀x
    Vậy A_(max)=4 khi x+1=0⇔x=-1
    B=2x-3x²-4
    =-3(x²-2/3x+4/3)
    =-3(x²-2/3x+1/9+11/9)
    =-3(x²-2/3x+1/9)-11/3
    =-3[x²-2.x. 1/3+(1/3)^2]-11/3
    =-3(x-1/3)^2-11/3
    Ta có:(x-1/3)^2≥0∀x
    ⇒3(x-1/3)^2≥0∀x
    ⇒-3(x-1/3)^2≤0∀x
    ⇒-3(x-1/3)^2-11/3≤-11/3∀x
    Vậy B_(max)=-11/3 khi x-1/3=0⇔x=1/3

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

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