b) – 5x -10 . (-7) = 55 c) |x+101| – (-16) = (-43).(-5) d) 3. |x| < 15 e) (x-7) . (x+1) = 0 f) (x-7) . (x+1) > 0

0 bình luận về “b) – 5x -10 . (-7) = 55 c) |x+101| – (-16) = (-43).(-5) d) 3. |x| < 15 e) (x-7) . (x+1) = 0 f) (x-7) . (x+1) > 0”

1. $b) – 5x -10 . (-7) = 55$

   $⇔-5x+70=55$

   $⇔-5x=55-70$

   $⇔-5x=-15$

   $⇔x=3$

   $c) |x+101| – (-16) = (-43).(-5)$

   $⇔ |x+101|+16=215$

   $⇔|x+101|=215-16$

   $⇔|x+101|=199$

   ⇒\(\left[ \begin{array}{l}x+101=199\\x+101=-199\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=199-101\\x==-199-101\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x=98\\x=-300\end{array} \right.\)

   $d)3. |x| < 15$

   $⇔|x|<5$

   ⇒\(\left[ \begin{array}{l}x<5\\x>-5\end{array} \right.\)

   $Vậy$ $-5<x<5$

   e) (x-7) . (x+1) = 0

   ⇒\(\left[ \begin{array}{l}x-7=0\\x+1=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\)

   $Vậy$ $x=7;x=-1$

   $f) (x-7) . (x+1) > 0$

   ⇒\(\left[ \begin{array}{l}\left \{ {{x-7>0} \atop {x+1>0}} \right.\\\left \{ {{x-7<0} \atop {x+1<0}} \right.\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}\left \{ {{x>7} \atop {x>-1}} \right.\\\left \{ {{x<7} \atop {x<-1}} \right.\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x>7\\x<-1\end{array} \right.\) 

   $Vậy$ $x>7$ $hoặc$ $x<-1$

    

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2. Đáp án:

    

   Giải thích các bước giải:

   b) – 5x  – 10 . (-7) = 55

   ⇔- 5x  + 70= 55

   ⇔- 5x         = 55 – 70

   ⇔- 5x         = -15

   ⇔    x         = -15 : -5

   ⇔    x         = 3

   Vậy x=3

   c) |x + 101| – (-16) = (-43) . (-5)

   ⇔|x + 101| + 16 = 215

   ⇔|x + 101|         = 215 – 16

   ⇔|x + 101|         = 199

   ⇔\(\left[ \begin{array}{l}x + 101= 199\\x + 101= -199\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x = 199-101\\x=-199 – 101\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x = 98\\x=-300\end{array} \right.\) 

   Vậy x ∈ {98;-300}

   d) 3. |x| < 15

   ⇔|x| < 15 : 3

   ⇔|x| < 5 

   ⇔-5<x<5

   e) (x-7) . (x+1) = 0

   ⇔\(\left[ \begin{array}{l}x -7= 0\\x + 1= 0\end{array} \right.\)

   ⇔\(\left[ \begin{array}{l}x = 7\\x=- 1\end{array} \right.\)

   Vậy x ∈ {7;-1}

   f) (x-7) . (x+1) > 0

   ⇔(x-7) và (x+1) cùng dấu

   TH1:

   $\left \{ {{x-7>0} \atop {x+1>0}} \right.$

   ⇔$\left \{ {{x>7} \atop {x>-1}} \right.$ 

   ⇒x>7

   TH2:

   $\left \{ {{x-7<0} \atop {x+1<0}} \right.$

   ⇔$\left \{ {{x<7} \atop {x<-1}} \right.$ 

   ⇒x<-1

   Vậy (x-7) . (x+1) > 0 khi \(\left[ \begin{array}{l}x>7\\x<-1\end{array} \right.\) 

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