Tìm x biết : a) (19 |x-1| + 2 . 5^2) : 14=(13-8)^2 -4^2 b) |39/2 – 3 . x^2|=15/2 c)||x+3|-8|=20 d) |3x-2^4|. 7^2019=2.7^2020.1/2019^0

0 bình luận về “Tìm x biết : a) (19 |x-1| + 2 . 5^2) : 14=(13-8)^2 -4^2 b) |39/2 – 3 . x^2|=15/2 c)||x+3|-8|=20 d) |3x-2^4|. 7^2019=2.7^2020.1/2019^0”

1. `a)`

   `( 19|x-1| +2 *5^2) : 14  = (13-8)^2 -4^2`

   ` => (19|x+1| + 2*25) : 14 = 5^2 -4^2`

   ` => ( 19 |x+1| + 50) : 14 = 25 – 16 = 9`

   ` => 19|x+1| + 50 = 9 *14`

   ` => 19|x+1| = 126 – 50 = 76`

   ` => |x+1| = 4`

   Trường hợp `1`

   ` x+ 1 = 4 => x= 3`

   Th 2

   ` x +1 = -4 => x= -4-1 = -5`

   Vậy ` x \in {3;-5}`

   `b)`

   ` |39/2 – 3x^2| = 15/2`

   TH1

   ` 39/2 -3x^2 = 15/2`

   ` => 3x^2 = 39/2 -15/2`

   ` => 3x^2 = 24/2 = 12`

   ` => x^2 = 4`

   ` => x= ±2`

   TH2

   ` 39/2 -3x^2 = -15/2`

   ` => 3x^2 – 39/2 +15/2 = 54/2 = 27`

   `=> x^2 = 9`

   ` => x=±3`

   Vậy ` x \in { -3 ; -2 ; 2 ; 3}`

   `c)`

   Th1

   ` |x+3| – 8 = 20`

   ` => |x+3| = 28`

   ` => x +3 = 28` hoặc ` x +3 = -28`

   ` => x= 25 ` hoặc ` x = -31`

   TH2 

   ` |x+3| -8 = -20`

   ` =>|x+3| = -12` ( vô lí vì ` |x+3| \ge 0` )

   Vậy ` x\in {25;-31}`

   `d)`

   ` |3x -2^4 | * 7^(2019) = 2* 7^(2020) * 1/(2019^0)`

   ` => |3x -2^4 | * 7^(2019) = 2* 7^(2020) * 1/1 `

   ` => |3x -2^4 | * 7^(2019) = 2* 7^(2020) `

   ` => |3x -2^4| = 2*7 = 14`

   ` => |3x-16| = 14`

   TH1

   ` 3x-16 = 14 => 3x = 30 => x= 10`

   TH2

   ` 3x -16 = -14 => 3x = 2 => x= 2/3`

   Vậy ` x\in {2/3 ; 10}`

    

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2. Đáp án:

   Giải thích các bước giải:

   `a) (19 |x-1| + 2 . 5^2) : 14=(13-8)^2 -4^2`

   `->(19|x-1|+50):14=5^2-4^2`

   `->(19|x-1|+50):14=9`

   `->19|x-1|+50=126`

   `->19|x-1|=76`

   `->|x-1|=4`

   `->` \(\left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\)

   Vậy `x∈{5;-3}`

   `b)` `|39/2 – 3 . x^2|=15/2`

   `->` `|39/2 – 3 . x^2|=15/2`\(\left[ \begin{array}{l}\dfrac{39}{2}-3x^2=\dfrac{15}{2}\\\dfrac{39}{2}-3x^2=-\dfrac{15}{2}\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}3x^2=12\\3x^2=27\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}x^2=4\\x^2=9\end{array} \right.\) 

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

   Vậy `x∈{±2;±3}`

   `c)` `|x+3|-8=20`

   `->|x+3|=28`

   `->` \(\left[ \begin{array}{l}x+3=28\\x+3=-28\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}x=25\\x=-31\end{array} \right.\) 

   Vậy `x∈{25;-31}`

   `d)` `|3x-2^4|. 7^(2019)=2.7^2020. 1/(2019^0)`

   `->|3x-16|.7^(2019)=2.7^(2020)`

   `->|3x-16|.7^(2019)=2.7^(2019) .7`

   `->|3x-16|=14`

   `->` \(\left[ \begin{array}{l}3x-16=14\\3x-16=-14\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}x=10\\x=\dfrac{2}{3}\end{array} \right.\) 

   Vậy `x∈{10;2/3}`

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