a,|x-(-5)|=2020 b,(-5)chia hết cho(x-2) c,6 chia hết cho(x+1) d,(3x-2)chia hết(x+1) Ai giúp mik với

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1. a,|x-(-5)|=2020

   ⇒|x+5|=2020

   ⇒\(\left[ \begin{array}{l}x+5=2020\\x+5=-2020\end{array} \right.\) =>\(\left[ \begin{array}{l}x=2015\\x=-2025\end{array} \right.\) 

   Vậy x=2015 hoặc x=-2015

   b, Ta có: -5$\vdots$x-2

   ⇒x-2∈Ư(5)={±1;±5}

   x-2=1⇒x=3

   x-2=-1⇒x=1

   x-2=5⇒x=7

   x-2=-5⇒x=-3

   Vậy x∈{3;1;7;-3}

   c, Ta có: 6$\vdots$x+1

   ⇒x+1∈Ư(6)={±1;±2;±3;±6}

   x+1=1⇒x=0

   x+1=-1⇒x=-2

   x+1=-2⇒x=-3

   x+1=2⇒x=1

   x+1=3⇒x=2

   x+1=-3⇒x=-4

   x+1=-6⇒x=-7

   x+1=6⇒x=5

   Vậy x∈{0;-2;-3;1;2;-4;-7;5}

   d, Ta có: 3x-2$\vdots$x+1

   ⇒3(x+1)-5$\vdots$x+1

   ⇒x+1∈Ư(5)={±1;±5}

   x+1=1⇒x=0

   x+1=-1⇒x=-2

   x+1=5⇒x=4

   x+1=-5⇒x=-6

   Vậy x∈{0;-2;4;-6}

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2. a/ | x – (-5) | = 2020

   ⇒ | x + 5 | = 2020

   ⇒ \(\left[ \begin{array}{l}x+5=2020\\x+5=-2020\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=2015\\x=-2015\end{array} \right.\)

   Vậy…

   —

   b/ (-5) ⋮ ( x – 2 )

   ⇒ ( x – 2 ) ∈ Ư(-5)

   ⇒ ( x – 2 ) ∈ { ±1; ±5 }

   ⇒ x ∈ { ±3; 1; 7 }

   Vậy…

   —

   c/ 6 ⋮ ( x + 1 )

   ⇒ ( x + 1 ) ∈ Ư(6)

   ⇒ ( x + 1 ) ∈ { ±1; ±2; ±3; ± 6 }

   ⇒ x ∈ { 0; ±2; -3; 1; -4; -7; 5 }

   Vậy…

   —

   d/ ( 3x – 2 ) ⋮ ( x + 1 )

   ⇒ [ 3( x + 1 ) – 5 ] ⋮ ( x + 1 )

   Vì 3( x + 1 ) ⋮ ( x + 1 ) nên để [ 3( x + 1 ) – 5 ] ⋮ ( x + 1 ) thì 5 ⋮ ( x + 1 )

   ⇒ ( x + 1 ) ∈ Ư(5)

   ⇒ ( x + 1 ) ∈ { ±1; ±5 }

   ⇒ x ∈ { 0; -2; 4; -6 }

   Vậy…

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