bài 1 tinh nhanh 407-{-512+[386+(407+512-386)]} -151+{-872-[361-(872+361+151)]} bài 2 tìm x thuoc z a)102+3(x-71)=-12-39 b)41-196:(2x-1)=25-33 c)

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1. Bài $1$.

   `407-{-512+[386+(407+512-386)]}`

   `= 407 – {-512 + [386 + 407 + 512 – 386]}`

   `= 407 – {-512 + 386 + 407 + 512 – 386}`

   $= 407 + 512 – 386 – 407 – 512 + 386$

   $= (407 – 407) + (512 – 512) + (-386+386)$

   $= 0$

   `-151+{-872-[361-(872+361+151)]}`

   `= -151 + {-872 – [361 – 872 – 361 – 151]}`

   `= -151 + {-872 – 361 + 872 + 361 + 151}`

   `= -151 + {(-872+872)+(-361+361) + 151}`

   $= -151 + 151$

   $= 0$

   Bài $2$.

   $a$) $102 + 3.(x-71) = -12-39$

   $⇔ 102 + 3.(x-71) = -51$

   $⇔ 3(x-71) = -153$

   $⇔ x – 71 = -51$

   $⇔ x = 20$

      Vậy $x=20$

   $b$) $41-196:(2x-1)=25-33$

   $⇔ 41 – 196:(2x-1) = -8$

   $⇔ 196:(2x-1) = 49$

   $⇔ 2x-1 = 4$

   $⇔ x = \dfrac{5}{2}$ ($KTM$)

     Vậy $x$ $∈$ $∅$

   $c)57-2.|31-x|=13$

   $⇔ 2.|31-x| = 44$

   $⇔ |31-x| = 22$

   $⇒$ \(\left[ \begin{array}{l}x=9\\x=53\end{array} \right.\) 

      Vậy $x$ $∈$ `{9;53}`

   $d)13-7.(3x+1)^2=-162$

   $⇔ 7.(3x+1)^2 = 175$

   $⇔ (3x+1)^2 = 25$

   $⇒$ $3x+1 = ±5$

   $⇒$ \(\left[ \begin{array}{l}x=\dfrac{4}{3}(KTM)\\x=-2\end{array} \right.\)

      Vậy $x=-2$

   $e)-31-324:(x+1)^3=-19$

   $⇔ 324 : (x+1)^3 = -12$

   $⇔ (x+1)^3 = -27$

   $⇔ x + 1 = -3$

   $⇔ x = -4$

      Vậy $x=-4$

   Bài $3$.

   $a$) $2x+81 \vdots x-1$

   $⇔ 2x+81 – 2(x-1) \vdots x-1$

   $⇔ 2x + 81 – 2x + 2 \vdots x-1$

   $⇔ 83 \vdots x-1$

   $⇒$ $x-1$ $∈$ `Ư(83)={±1;±83}`

   $⇒$ $x$ $∈$ `{-82;0;2;84}`

     Vậy $x$ $∈$ `{-82;0;2;84}`

   $b$) $3x-36  \vdots x+1$

   $⇔ 3x – 36 – 3(x+1) \vdots x+1$

   $⇔ 3x – 36 – 3x – 3 \vdots x+1$

   $⇔ -39 \vdots x+1$

   $⇒$ $x+1$ $∈$ `Ư(39)={±1;±3;±13;±39}`

   $⇒$ $x$ $∈$ `{-40;-14;-4;-2;0;2;12;38}`

     Vậy $x$ $∈$ `{-40;-14;-4;-2;0;2;12;38}`

   $c$) $5x-29 \vdots x-1$

   $⇔ 5x-29 – 5(x-1) \vdots x-1$

   $⇔ 5x-29 – 5x + 5 \vdots x-1$

   $⇔ -24 \vdots x-1$

   $⇒$ $x-1$ $∈$ `Ư(24)={±1;±2;±3;±4;±6;±8;±12;±24}`

   $⇒$ $x$ $∈$ `{-23;-11;-7;-5;-3;-2;-1;0;2;3;4;5;7;9;13;25}`

     Vậy $x$ $∈$ `{-23;-11;-7;-5;-3;-2;-1;0;2;3;4;5;7;9;13;25}`

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