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

Toán Lớp 6: a, (3x+8)chia hết (x+4) b, (15x+3)chia hết (4x+5) c, (15x+3) chia hết ( 3x+1) d, (10x+6) chia hết (2x+3)

Toán Lớp 6: a, (3x+8)chia hết (x+4)
b, (15x+3)chia hết (4x+5)
c, (15x+3) chia hết ( 3x+1)
d, (10x+6) chia hết (2x+3)

Comments ( 2 )

  1. a,
    => 3 x + 8 + ( x + 4 )   ⋮ ( x + 4 )
    => 3 x + 8 + 3 ( x + 4 ) ⋮ ( x + 4 ) 
    => 3 x + 8 + ( 3 x + 12 ) ⋮ ( x + 4 ) 
    => 12 ⋮ ( x + 4 )  ⇒ ( x + 4 ) ∈ Ư ( 12 ) 
    => Ư ( 12 ) = { 1 ; 2 ; 3 ; 4 ; 6 ; 12 }
    => x ∈ { − 3 ; − 2 ; − 1 ; 0 ; 2 ; 8 } 
    Vậy : x ∈ { − 3 ; − 2 ; − 1 ; 0 ; 2 ; 8 } 
    b,  
    => 15 x + 3 + 4 ( x + 5 ) ⋮ ( 4 x + 5 ) 
    => 15 x + 3 + ( 4 x + 20 ) ⋮ ( 4 x + 5 ) 
    => 20 ⋮ ( 4 x + 5 )  ⇒ ( 4 x + 5 ) ∈ Ư ( 20 ) 
    => Ư ( 20 ) = { 1 ; 2 ; 4 ; 5 ; 10 ; 20 } 
    => ( 4 x + 5 ) = { 1 ; 2 ; 4 ; 5 ; 10 ; 20 } 
    => x = { − 16 ; − 12 ; − 4 ; 0 ; 20 ; 60 } 
    Vậy : x = { − 16 ; − 12 ; − 4 ; 0 ; 20 ; 60 } 
    c ,
    => 15 x + 3 + 3 ( x + 1 ) ⋮ ( 3 x + 1 ) 
    => 15 x + 3 + ( 3 x + 3 ) ⋮ ( 3 x + 1 ) 
    => 3 ⋮ ( 3 x + 1 )  
    =>( 3 x + 1 ) ∈ Ư ( 3 ) 
    => Ư ( 3 ) = { 1 ; 3 }  
    => ( 3 x + 1 ) = { 1 ; 3 } 
    => x = { 0 ; 1 , 5 } 
    Vậy : x = { 0 ; 1 , 5 }
    d , ( 10 x + 6 ) ⋮ ( 2 x + 3 ) 
    => 10 x + 6 + 2 ( x + 3 ) ⋮ ( 2 x + 3 ) 
    => 10 x + 6 + ( 2 x + 6 ) ⋮ ( 2 x + 3 ) 
    => ( 2 x + 3 ) ∈ Ư ( 6 ) 
    => Ư ( 6 ) = { 1 ; 2 ; 3 ; 6 }  
    => ( 2 x + 3 ) ∈ { 1 ; 2 ; 3 ; 6 }  
    => x ∈ { − 4 ; − 2 ; 0 ; 1 , 5 }
    Vậy : x ∈ { − 4 ; − 2 ; 0 ; 1 , 5 }
     

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a, (3x + 8) \vdots (x + 4) 
    => 3x + 8 + ( x + 4)  \vdots (x + 4)
    => 3x + 8 + 3(x + 4) \vdots (x + 4)
    => 3x + 8 + (3x + 12) \vdots (x + 4)
    => 12 \vdots (x + 4)
    => (x + 4) in Ư(12)
    => Ư(12) = {1 ; 2 ; 3 ; 4 ; 6 ; 12}
    => x in {-3 ; -2 ; -1 ; 0 ; 2 ; 8}
    b, (15x+3) \vdots (4x+5)
    => 15x + 3 + 4(x + 5) \vdots (4x + 5)
    => 15x + 3 + (4x + 20) \vdots (4x + 5)
    => 20 \vdots (4x + 5)
    => (4x + 5) in Ư(20)
    => Ư(20) = {1 ; 2 ; 4 ; 5 ; 10 ; 20}
    => (4x + 5) = {1 ; 2 ; 4 ; 5 ; 10 ; 20}
    => x = {-16; -12 ; -4 ; 0 ; 20 ; 60}
    c, (15x+3) \vdots ( 3x+1)
    => 15x + 3 + 3(x + 1) \vdots ( 3x+1)
    => 15x + 3 + (3x + 3) \vdots ( 3x+1)
    => 3 \vdots ( 3x+1)
    => (3x+1) in Ư(3)
    => Ư(3) = {1;3}
    => (3x+1) = {1;3}
    => x = {0;1,5}
    d,(10x+6) \vdots (2x+3)
    => 10x + 6 + 2(x+3) \vdots (2x+3)
    => 10x + 6 + (2x+6) \vdots (2x+3)
    => (2x+3) in Ư(6)
    => Ư(6) = {1;2;3;6}
    => (2x+3) in {1;2;3;6}
    => x in {-4;-2;0;1,5}

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

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