Bài 1: Tìm x ∈ Z biết:
20) – 12 ( x – 5 ) + 7 ( 3 – x ) = 5
21) ( x – 2 ) . ( x + 4 ) = 0
22) ( x – 2 ) . ( x + 15 ) = 0
23) ( 7 – x ) . ( x + 19 ) = 0
24) – 5 < x < 1
25) | x | < 3
26) ( x - 3 ) ( x - 5 ) = 0
27) ( x - 5 ) ( x - 7) = 0
28) 2x ² - 3 = 29
29 - 6x - ( -7 ) = 25
30) 46 - ( x - 11 ) = - 48
31) | x - 12 | - 11 = 9
33) | x + 19 | + 24 = 23
34) ( x - 11 ) . ( x + 5 ) = 0
36) | x + 1 | - 5 = 10
Các bn giúp mk vs, mk dg cần gấp
Bài 1: Tìm x ∈ Z biết: 20) – 12 ( x – 5 ) + 7 ( 3 – x ) = 5 21) ( x – 2 ) . ( x + 4 ) = 0 22) ( x – 2 ) . ( x + 15 ) = 0 23) ( 7 – x ) . ( x + 19 ) =
By Aaliyah
Đáp án:
36) \(\left[ \begin{array}{l}
x = – 16\\
x = 14
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
20) – 12\left( {x – 5} \right) + 7\left( {3 – x} \right){\rm{ = }}5\\
\to – 12x + 60 + 21 – 7x = 5\\
\to 19x = – 76\\
\to x = – 4\\
21)\left( {x – 2} \right).\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x – 2 = 0\\
x + 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 4
\end{array} \right.\\
22)\left( {x – 2} \right)\left( {x + 15} \right) = 0\\
\to \left[ \begin{array}{l}
x – 2 = 0\\
x + 15 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 15
\end{array} \right.\\
23)\left( {7 – x} \right)\left( {x + 19} \right) = 0\\
\to \left[ \begin{array}{l}
7 – x = 0\\
x + 19 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = – 19
\end{array} \right.\\
24) – 5 < x < 1\\
\to x \in \left\{ { – 4; – 3; – 2; – 1;0} \right\}\\
25)\left| x \right| < 3\\
\to – 3 < x < 3\\
\to x \in \left\{ { – 2; – 1;0;1;2} \right\}\\
26)\left( {x – 3} \right)\left( {x – 5} \right) = 0\\
\to \left[ \begin{array}{l}
x – 3 = 0\\
x – 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 5
\end{array} \right.\\
27)\left( {x – 5} \right)\left( {x – 7} \right) = 0\\
\to \left[ \begin{array}{l}
x – 5 = 0\\
x – 7 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 7
\end{array} \right.\\
29) – 6x + 7 = 25\\
\to – 6x = 18\\
\to x = – 3\\
30)46 – x + 11 = – 48\\
\to x = 105\\
31)\left| {x – 12} \right| = 20\\
\to \left[ \begin{array}{l}
x – 12 = 20\\
x – 12 = – 20
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 32\\
x = – 8
\end{array} \right.\\
34)\left( {x – 11} \right)\left( {x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 11\\
x = – 5
\end{array} \right.\\
36)\left| {x + 1} \right| = 15\\
\to \left[ \begin{array}{l}
x + 1 = – 15\\
x + 1 = 15
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 16\\
x = 14
\end{array} \right.
\end{array}\)