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

Toán Lớp 7: 2a+b+c+d/a=a+2b+c+d/b=a+b+2c+d/c=a+b+c+2d/d Tính M biết :M=a+b/c+d+b+c/d+a+c+d/a+b+d+a/b+c giúp mik nha

Toán Lớp 7: 2a+b+c+d/a=a+2b+c+d/b=a+b+2c+d/c=a+b+c+2d/d
Tính M biết :M=a+b/c+d+b+c/d+a+c+d/a+b+d+a/b+c
giúp mik nha

Comments ( 2 )

  1. Giải đáp:
    $M = 4$
    Lời giải và giải thích chi tiết:
    Áp dụng tính chất dãy tỉ số bằng nhau :
    $\frac{2a+b+c+d}{a} = \frac{a+2b+c+d}{b} = \frac{a+b+2c+d}{c} = \frac{a+b+c+2d}{d} = \frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}$
    ⇔ $\frac{2a+b+c+d}{a} = \frac{a+2b+c+d}{b} = \frac{a+b+2c+d}{c} = \frac{a+b+c+2d}{d} = \frac{5a+5b+5c+5d}{a+b+c+d} = 5$
    ⇔ $2a + b + c + d = 5a , a + 2b + c + d = 5b , a + b + 2c + d = 5c , a + b + c + 2d = 5d$
    ⇔ $b + c + d = 3a , a + c + d = 3b , a + b + d = 3c , a + b + c = 3d$
    Ta có : $3a + 3b = b + c + d + a + c + d$
    ⇔ $3a + 3b = a + b + 2c + 2d$
    ⇔ $2a + 2b = 2c + 2d$
    ⇔ $a + b = c + d$
    ⇔ $\frac{a+b}{c+d} = 1$
    Chứng minh tương tự :
    $\frac{b+c}{d+a} = 1 , \frac{c+d}{a+b} = 1 , \frac{d+a}{b+c} = 1$
    Lại có :
    $M = \frac{a+b}{c+d} + \frac{b+c}{d+a} + \frac{c+d}{a+b} + \frac{d+a}{a+b}$
    ⇔ $M = 1 + 1 + 1 + 1$
    ⇔ $M = 4$

  2. text{Ta có :}
    $\dfrac{2a + b + c + d}{a}$ = $\dfrac{a + 2b + c + d}{b}$
    = $\dfrac{a + b + 2c + d}{c}$ = $\dfrac{a + b + c + 2d}{d}$ 
    => $\dfrac{a + b + c + d}{a}$ + 1 = $\dfrac{a + b + c + d}{b}$ + 1
    = $\dfrac{a + b + c + d}{c}$ + 1 = $\dfrac{a + b + c + d}{d}$ + 1
    => $\dfrac{a + b + c + d}{a}$ = $\dfrac{a + b + c + d}{b}$
    = $\dfrac{a + b + c + d}{c}$ = $\dfrac{a + b + c + d}{d}$ ( 1 )
    text{Nếu} a + b + c + d = 0
    => $\left[\begin{matrix} a + b = -(c + d)\\ b + c = -(a + d)\\c + d = -(a + b)\\a + d = -(b + c )\end{matrix}\right.$
    text{Ta có}
    M = $\dfrac{a+b}{c+d}$ + $\dfrac{b+c}{d+a}$ + $\dfrac{c+d}{a+b}$ + $\dfrac{a+d}{b+c}$
    => M = $\dfrac{-(c+d)}{c+d}$ + $\dfrac{-(a+d)}{d+a}$ + $\dfrac{-(a+b)}{a+b}$ + $\dfrac{-(b+c)}{b+c}$
    => M = (-1) + (-1) + (-1) + (-1)
    => M = -4
    text{Nếu} a + b + c + d  $\ne$ 0
    => a = b = c = d
    => M = $\dfrac{a+b}{c+d}$ + $\dfrac{b+c}{d+a}$ + $\dfrac{c+d}{a+b}$ + $\dfrac{a+d}{b+c}$
    = 1 + 1 + 1 + 1
    => M = 4
    text{Vậy} : M ∈ { 4 ; -4 }

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

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