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

Toán Lớp 8: Tìm GTNN của các biểu thức sau a, M=x ²+3x-4 b, N=3x ²-2x-4 c, P= x ²+y ²-2x-4y+8

Toán Lớp 8: Tìm GTNN của các biểu thức sau
a, M=x ²+3x-4
b, N=3x ²-2x-4
c, P= x ²+y ²-2x-4y+8

Comments ( 2 )

  1. a)
    M=x^2+3x-4
    =x^2+2.x. 3/2+9/4-4-9/4
    =(x+3/2)^2-25/4>=-25/4
    Dấu “=” xảy ra khi : (x+3/2)^2=0
    <=> x=-3/2
    Vậy M_(min)=-25/4 <=> x=-3/2
    b)
    N=3x^2-2x-4
    =3(x^2-2/3x-4/3)
    =3(x^2-2.x . 1/3+1/9-4/3-1/9)
    =3[(x-1/3)^2-13/9]
    =3(x-1/3)^2-13/3>=-13/3
    Dấu “=” xảy ra khi : (x-1/3)^2=0
    <=> x=1/3
    Vậy N_(min)=-13/3 <=> x=1/3
    c)
    P=x^2+y^2-2x-4y+8
    =(x^2-2x+1)+(y^2-4y+4)+3
    =(x-1)^2+(y-2)^2+3>=3
    Dấu “=” xảy ra khi : {((x-1)^2=0),((y-2)^2=0):} <=> {(x=1),(y=2):}
    Vậy P_(min)=3 <=> (x;y)=(1;2)
     

  2. a, $M=x^2+3x-4$
    $M=x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}$
    $M=(x+\dfrac{3}{2})^2-\dfrac{25}{4}$
    $(x+\dfrac{3}{2})^2≥0∀x$
    $⇔(x+\dfrac{3}{2})^2-\dfrac{25}{4}≥-\dfrac{25}{4}∀x$
    Dấu $=$ xảy ra khi 
    $x+\dfrac{3}{2}=0$
    $⇔x=\dfrac{-3}{2}$
    Vậy $M_{min}=\dfrac{-25}{4}⇔x=\dfrac{3}{2}$
    ————————————
    b, $N=3x^2-2x-4$
    $N=3x^2-2.\sqrt{3}x.\dfrac{\sqrt{3}}{3}+\dfrac{1}{3}-\dfrac{13}{3}$
    $N=(\sqrt{3}x-\dfrac{\sqrt{3}}{3})^2-\dfrac{13}{3}$
    $(\sqrt{3}x-\dfrac{\sqrt{3}}{3})^2≥0∀$
    $⇔(\sqrt{3}x-\dfrac{\sqrt{3}}{3})^2-\dfrac{13}{3}≥\dfrac{-13}{3}∀x$
    Dấu $=$ xảy ra khi
    $\sqrt{3}x-\dfrac{\sqrt{3}}{3}=0$
    $⇔\sqrt{3}x=\dfrac{\sqrt{3}}{3}$
    $⇔x=\dfrac{1}{3}$
    Vậy $N_{min}=\dfrac{-13}{3}⇔x=\dfrac{1}{3}$
    —————————————–
    c, $P=x^2+y^2-2x-4y+8$
    $P=x^2-2x+1+y^2-4y+4+3$
    $P=(x-1)^2+(y-2)^2+3$
    $(x-1)^2≥∀x,(y-2)^2≥0∀y$
    $⇔(x-1)^2+(y-2)^2+3≥3∀x,y$
    Dấu $”=”$ xảy ra khi
    $+,x-1=0$
    $⇔x=1$
    $+,y-2=0$
    $⇔y=2$
    Vậy $P_{min}=3⇔x=1;y=2$

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

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