a) [ 124 – ( 20 – 4x ) ] : 30 + 7 = 11 b) 3 ( x + 7 ) + 29 = 9 – 2 ( x + 5 ) c) ( x -trừ 1/2 ) ( -3 trừ x/2 = 0 )

0 bình luận về “a) [ 124 – ( 20 – 4x ) ] : 30 + 7 = 11 b) 3 ( x + 7 ) + 29 = 9 – 2 ( x + 5 ) c) ( x -trừ 1/2 ) ( -3 trừ x/2 = 0 )”

1. Đáp án:

   `a,[124-(20-4x)]:30+7=11`

   `->[124-20+4x]:30=4`

   `->104+4x=120`

   `->4x=16`

   `->x=4`

   Vậy `x=4`

   `b,3(x+7)+29=9-2(x+5)`

   `->3x+21+29=9-2x-10`

   `->3x+50=-1-2x`

   `->5x=-51`

   `->x=-51/5`

   Vậy `x=-51/5`

   `c,(x-1/2)((-3-x)/2)=0`

   `->` \(\left[ \begin{array}{l}x-\dfrac{1}{2}=0\\-3-x=0\end{array} \right.\)

   `->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-3\end{array} \right.\) 

   Vậy `x\in{-3;1/2}`

    

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2. a)[124-(20-4x)]:30+7=11

   =>[124-(20-4x)]:30=11-7

   => [124-(20-4x)]:30=4

   =>[124-(20-4x)]=4.30

   =>20-4x=244

   =>4x=244+20

   =>x=264:4

   =>x=66

   b)3(x+7)+29=9-2(x+5)

   =>3(x+7)+2(x+5)=9-29

   =>5(x+7)+(x+5)=-20

   =>(x+7)+(x+5)=-20:5

   => (x+7)+(x+5)=-4

   c)(x-1/2)(-3-x/2)=0

   => (x-1/2)=0 hoặc -3-x/2=0

   -với x-1/2=0=>x=1/2

   -với -3-x/2=0=>x=(-6)

    

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