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

Toán Lớp 8: Bài 1: Tìm GTNN N= 2x^2+4y^2-2x-4y+15 Bài 2:Tìm X a) (2x+3)^2-25(1-x)^2=0

Toán Lớp 8: Bài 1: Tìm GTNN
N= 2x^2+4y^2-2x-4y+15
Bài 2:Tìm X
a) (2x+3)^2-25(1-x)^2=0

Comments ( 2 )

  1. Giải đáp:
    Bài 1 ;
    GTNN $N = \frac{27}{2}$ khi $x = y = \frac{1}{2}$
    Bài 2 :
    \(\left[ \begin{array}{l}x=\frac{8}{3}\\x=\frac{2}{7}\end{array} \right.\) 
    Lời giải và giải thích chi tiết:
    Bài 1 :
    $N = 2x^{2} + 4y^{2} – 2x – 4y + 15$
    $N = 2( x^{2} – x + \frac{1}{4} ) + ( 4y^{2} – 4y + 1 ) + \frac{27}{2}$
    $N = 2( x – \frac{1}{2} )^{2} + ( 2y – 1 )^{2} + \frac{27}{2}$
    Vì $2( x – \frac{1}{2} )^{2} ≥ 0 , ( 2y – 1 )^{2} ≥ 0$ với $∀ x , y$
    ⇒ $2( x – \frac{1}{2} )^{2} + ( 2y – 1 )^{2} ≥ 0$
    ⇒ $2( x – \frac{1}{2} )^{2} + ( 2y – 1 )^{2} + \frac{27}{2} ≥ \frac{27}{2}$
    ⇔ $N ≥ \frac{27}{2}$
    Dấu “=” xảy ra ⇔ $x – \frac{1}{2} = 0 , 2y – 1 = 0$
    ⇔ $x = y = \frac{1}{2}$
    Bài 2 :
    $( 2x + 3 )^{2} – 25( 1 – x )^{2} = 0$
    ⇔ $[ ( 2x + 3 ) + 5( 1 – x ) ][ ( 2x + 3 ) – 5( 1 – x ) ] = 0$
    ⇔ $( 2x + 3 + 5 – 5x )( 2x + 3 – 5 + 5x ) = 0$
    ⇔ $( 8 – 3x )( 7x – 2 ) = 0$
    ⇔ \(\left[ \begin{array}{l}x=\frac{8}{3}\\x=\frac{2}{7}\end{array} \right.\) 

  2. $\\$
    Bài 1.
    N = 2x^2 +4y^2 – 2x – 4y +15
    -> N = (2x^2 –  2x) + (4y^2 – 4y)+15
    ->N = 2 (x^2  – x)+4 (y^2 – y) +15
    -> N = 2 (x^2 – 2 . x . 1/2 + 1/4) + 4 (y^2 – 2 . y  . 1/2 +1/4) + 27/2
    ->N  =2 (x-1/2)^2 + 4 (y-1/2)^2 + 27/2 ≥ 27/2 ∀x,y
    Dấu “=” xảy ra khi :
    ↔ (x-1/2)^2=0 ,(y-1/2)^2=0
    ↔x-1/2=0,y-1/2=0
    ↔x=y=1/2
    Vậy min N=27/2 ↔x=y=1/2
    $\\$
    Bài 2.
    (2x+3)^2 – 25 (1-x)^2=0
    -> (2x+3)^2-  5^2(1-x)^2=0
    -> (2x+3)^2 – (5 – 5x)^2 =0
    -> (2x+3-5+5x)(2x+3+5-5x)=0
    -> (7x – 2) (-3x + 8)=0
    TH1 : 
    7x-2=0
    -> 7x=2
    ->x=2/7
    TH2 :
    -3x+8=0
    ->-3x=-8
    ->x=8/3
    Vậy x=2/7 hoặc x=8/3
     

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