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

Toán Lớp 9: Tìm GTNN của a) x^2-4x+5 b) 2x^2+3x+5 c) 3x^2+6x+2 d) 3x^2-5x+1

Toán Lớp 9: Tìm GTNN của
a) x^2-4x+5
b) 2x^2+3x+5
c) 3x^2+6x+2
d) 3x^2-5x+1

Comments ( 2 )

  1. a. $x^2-4x+5$
    $= (x^2-4x+4)+1$
    $= (x-2)^2+1$
    Vì $(x-2)^2\ge 0 \; \forall x$
    $\Rightarrow (x-2)^2+1 \ge 1\;\forall x$
    Vậy $\min = 1$ khi $x-2=0 \Leftrightarrow x=2$
    b. $2x^2+3x+5$
    $= 2\left(x^2+\dfrac32x+\dfrac52\right)$
    $= 2\left(x^2+\dfrac32x+\dfrac9{16}+\dfrac{31}{16}\right)$
    $= 2\left[\left(x+\dfrac34\right)^2+\dfrac{31}{16}\right]$
    $= 2\left(x+\dfrac34\right)^2+\dfrac{31}8$
    Vì $2\left(x+\dfrac34\right)^2 \ge 0 \; \forall x$
    $\Rightarrow 2\left(x+\dfrac34\right)^2+\dfrac{31}8 \ge \dfrac{31}8\;\forall x$
    Vậy $\min =\dfrac{31}8$ khi $x+\dfrac34=0 \Leftrightarrow x=-\dfrac34$
    c. $3x^2+6x+2$
    $= 3(x^2+2x+\dfrac23)$
    $= 3(x^2+2x+1-\dfrac13)$
    $= 3(x+1)^2-1$
    Vì $3(x+1)^2\ge 0 \; \forall x$
    $\Rightarrow 3(x+1)^2-1 \ge -1\;\forall x$
    Vậy $\min =- 1$ khi $x+1=0 \Leftrightarrow x=-1$
    d. $3x^2-5x+1$
    $= 3\left(x^2-\dfrac{5}{3}x+25/36-\dfrac{13}{36}\right)$
    $= 3\left[\left(x-\dfrac56\right)^2-\dfrac{13}{36}\right]$
    $= 3\left(x-\dfrac56\right)^2-\dfrac{13}{12}$
    Vì $3\left(x-\dfrac56\right)^2 \ge 0 \; \forall x$
    $\Rightarrow 3\left(x-\dfrac56\right)^2-\dfrac{13}{12} \ge -\dfrac{13}{12}\;\forall x$
    Vậy $\min =-\dfrac{13}{12}$ khi $x-\dfrac56=0 \Leftrightarrow x=\dfrac56$

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a) x^2-4x+5
    =x^2-4x+4+1
    =(x-2)^2+1>=1
    Dấu “=” xảy ra khi (x-2)^2=0
    <=> x=2
    Vậy $GTNN_{x^2-4x+5}=1 \Leftrightarrow x=2$ 
    b) 2x^2+3x+5
    =2.(x^2+3/2x+5/2)
    =2.(x^2+2.x . 3/4+9/16+5/2-9/16)
    =2.[(x+3/4)^2+31/16]
    =2.(x+3/4)^2+31/8>=31/8
    Dấu “=” xảy ra khi (x+3/4)^2=0
    <=> x=-3/4
    Vậy $GTNN_{2x^2+3x+5}=\dfrac{31}{8} \Leftrightarrow x=-\dfrac{3}{4}$
    c) 3x^2+6x+2
    =3.(x^2+2x+2/3)
    =3.(x^2+2x+1+2/3-1)
    =3.[(x+1)^2-1/3]
    =3.(x+1)^2-1>=-1
    Dấu “=” xảy ra khi (x+1)^2=0
    <=> x=-1
    Vậy $GTNN_{3x^2+6x+2}=-1 \Leftrightarrow x=-1$
    d) 3x^2-5x+1
    =3.(x^2-5/3x+1/3)
    =3.(x^2-2.x . 5/6+25/36+1/3-25/36)
    =3.[(x-5/6)^2-13/36]
    =3.(x-5/6)^2-13/12>=-13/12
    Dấu “=” xảy ra khi (x-5/6)^2=0
    <=> x=5/6
    Vậy $GTNN_{3x^2-5x+1}=-\dfrac{13}{12} \Leftrightarrow x=\dfrac{5}{6}$

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

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