=>-(1+2+...+x)=-36

=>1+2+...+x=36

\(\Leftrightarrow x=8\)

-(1+2+...+x)=-36

=1+2+...+x=36

<-> x=8

a) {-2).x = -10 => x = (-10): (- 2) => x = 5.

b)  (-18). x = -36 => x = (-36): (-18) => x = 2.

c) 2.x + 1 = 3=>2x = 2 => x = l.

d) (-4).x + 5 = -15 => (-4)x = (-15) -5 => (-4) x = -20 => x = 5.

a) {-2).x = -10 => x = (-10): (- 2) => x = 5.

b)  (-18). x = -36 => x = (-36): (-18) => x = 2.

c) 2.x + 1 = 3=>2x = 2 => x = l.

d) (-4).x + 5 = -15 => (-4)x = (-15) -5 => (-4) x = -20 => x = 5.

1. x + 2x = -36

=> 3x = -36

=> x = -36 : 3

=> x = -12

2. (2x + 3) \(⋮\)(x - 2)

=> (2x - 2) + 5 \(⋮\)(x - 2)

=> 2(x - 2) + 5 \(⋮\)(x - 2)

=> 5 \(⋮\)(x - 2)

=> x - 2 \(\in\)Ư(5) = {-5;-1;1;5}

=> x \(\in\){-3;1;3;7}

3. Khi đó a . (-b) = -132

4. -2(3x + 2) = 1² + 2² + 3²

=> -2(3x + 2) = 1 + 4 + 9

=> -2(3x + 2) = 14

=> 3x + 2 = 14 : (-2)

=> 3x+ 2 = -7

=> 3x = -7 - 2

=> 3x = -9

=> x = -9 : 3

=> x = -3

1/ \(x+2x=-36\)

\(\Rightarrow3x=-36\)

\(\Rightarrow x=-\frac{36}{3}\)

\(\Rightarrow x=-12\)

2/    \(\left(2x+3\right)⋮\left(x-2\right)\)

\(\Leftrightarrow\left(2x-4\right)+7⋮\left(x-2\right)\)

\(\Leftrightarrow2\left(x-2\right)+7⋮\left(x-2\right)\)

\(\Rightarrow7⋮\left(x-2\right)\)

\(\Rightarrow\left(x-2\right)\inƯ\left(7\right)\)

\(\Rightarrow x\inƯ\left(7-2\right)\)

\(\Rightarrow x\inƯ\left(5\right)\)

\(\Rightarrow x\in\left\{-5,1,5\right\}\)

Vậy x nhỏ nhất để \(\left(2x-3\right)⋮\left(x-2\right)\) là -5

3/ Vì \(a\cdot b=32\)

\(\Rightarrow-a\cdot b=-\left(a\cdot b\right)=-32\)

4/ \(-2\left(3x+2\right)=1^2+2^2+3^2\)

\(\Leftrightarrow-6x-4=1+4+9\)

\(\Leftrightarrow-6x=14+4\)

\(\Leftrightarrow-6x=18\)

\(\Leftrightarrow x=\frac{18}{-6}\)

\(\Rightarrow x=3\)

a: =>-12<x<2y<-9

=>x=-11; y=-5

b: =>-7<3(x-1)<8

\(\Leftrightarrow3\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)

\(\Leftrightarrow x-1\in\left\{2;1;0;-1;-2\right\}\)

hay \(x\in\left\{3;2;1;0;-1\right\}\)

Lớp 4?

a) \(\dfrac{x+1}{4}=\dfrac{36}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=144\)

\(\Rightarrow\left[{}\begin{matrix}x+1=12\\x+1=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-13\end{matrix}\right.\)

Vậy: \(x\in\left\{11;-13\right\}\)

b) \(\dfrac{x}{7}=\dfrac{55x-4}{28}\)

\(\Rightarrow4x=55x-4\)

\(\Rightarrow-51x=-4\)

\(\Rightarrow x=\dfrac{4}{51}\)

Vậy: \(x=\dfrac{4}{51}\)

a) \(\dfrac{x + 1}{4} = \dfrac{36}{x + 1} \) 

\(\Rightarrow\) \(( x + 1 )( x + 1 ) = 36 . 4 \)  

\(\Rightarrow ( x + 1 )^2 = 144 \) 

\(\Rightarrow ( x + 1 )^2 = 12^2 = ( -12 )^2 \) 

\(\Rightarrow\) \(x + 1 ∈ \) { \(12 ; -12 \) }

\(\Rightarrow \) \(x \) \(∈ \) { \(11 ; -13 \) }

Vậy \(x ∈ \) { \(11 ; -13 \) } 

Bài làm:

Ta có: \(a^2.\left(a+1\right)=36\)

\(\Leftrightarrow a^3+a^2-36=0\)

\(\Leftrightarrow\left(a^3-3a^2\right)+\left(4a^2-12a\right)+\left(12a-36\right)=0\)

\(\Leftrightarrow a^2\left(a-3\right)+4a\left(a-3\right)+12\left(a-3\right)=0\)

\(\Leftrightarrow\left(a-3\right)\left(a^2+4a+12\right)=0\)

Mà \(a^2+4a+12=\left(a+2\right)^2+8>0\)

\(\Rightarrow a-3=0\Rightarrow a=3\)