Giải pt: a) 4x – 10 = 0 b) 7 – 3x = 9 – x c) 2x – (3 – 5x) = 4(x +3) d) 5 – (6 – x) = 4(3 – 2x) e) 4(x +3) = -7x +17 f) 5(x -3) = 2(x -1) + 7

0 bình luận về “Giải pt: a) 4x – 10 = 0 b) 7 – 3x = 9 – x c) 2x – (3 – 5x) = 4(x +3) d) 5 – (6 – x) = 4(3 – 2x) e) 4(x +3) = -7x +17 f) 5(x -3) = 2(x -1) + 7”

1. `a) 4x – 10 = 0`

   `⇔4x=10`

   `⇔x=5/2`

   Vậy `S={5/2}`

   `b) 7 – 3x = 9 – x`

   `⇔-3x+x=9-7`

   `⇔-2x=2`

   `⇔x=1`

   Vậy `S={1}`

   `c) 2x – (3 – 5x) = 4(x +3)`

   `⇔2x-3+5x=4x+12`

   `⇔2x+5x-4x=12+3`

   `⇔3x=15`

   `⇔x=5`

   Vậy `S={5}`

   `d) 5 – (6 – x) = 4(3 – 2x)`

   `⇔ 5 – 6 + x) = 12 – 8x)`

   `⇔x+8x=12-5 + 6`

   `⇔9x=13`

   `⇔x=13/9`

   Vậy `S={13/9}`

   `e) 4(x +3) = -7x +17`

   `⇔4x+12=-7x+17`

   `⇔4x+7x=17-12`

   `⇔11x=5`

   `⇔x=5/11`

   Vậy `S={5/11}`

   `f) 5(x -3) = 2(x -1) + 7`

   `⇔5x-15=2x-2+7`

   `⇔5x-2x=-2+7+15`

   `⇔3x=20`

   `⇔x=20/3`

   Vậy `S={20/3}`

    

   Trả lời

2. a) 4x – 10 = 0

   ⇔4x=10

   ⇔x=$\frac{5}{2}$ 

   Vậy S={$\frac{5}{2}$}

   b) 7 – 3x = 9 – x

   ⇔-3x+x=9-7

   ⇔-2x=2

   ⇔x=-1

   Vậy S={-1}

   c) 2x – (3 – 5x) = 4(x +3)

   ⇔2x-3+5x=4x+12

   ⇔2x+5x-4x=12+3

   ⇔3x=15

   ⇔x=5

   Vậy S={5}

   d) 5 – (6 – x) = 4(3 – 2x)

   ⇔5-6+x=12-8x

   ⇔x+8x=12-5+6

   ⇔9x=13

   ⇔x=$\frac{13}{9}$ 

   Vậy S={$\frac{13}{9}$}

   e) 4(x +3) = -7x +17

   ⇔4x+12=-7x+17

   ⇔4x+7x=17-12

   ⇔11x=5

   ⇔x=$\frac{5}{11}$

   Vậy S={$\frac{5}{11}$}

   f) 5(x -3) = 2(x -1) + 7

   ⇔5x-15=2x-2+7

   ⇔5x-2x=-2+7+15

   ⇔3x=20

   ⇔x=$\frac{20}{3}$

   Vậy S={$\frac{20}{3}$}

   Gửi????

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   @mind????

   #Dinosaurss

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