x.(4+x)

=4x+x²

tk nha!

15 - ( 13 + x) = x- ( 23 - 17 )

15-13-x=x-23+17

2x=(-15)+(-13)-23+17

2x=(-28)-23+17

2x=(-51)+17

2x=(-34)

x=(-34):2

x=(-17)

15 - ( 13 + x ) = x - ( 23 - 17 )

15 - 13 - x = x - 6 

-x - x = -6 + ( -2 )

-x + ( -x ) = -8

2.(-x) = -8

     -x = -8 : 2

     -x = -4

=> x = 4

Vậy x = 4

Chúc bạn học giỏi

____Ngốc cuồng BTS____:

-75+(-49)+|-75|+49-12

=-75+(-49)+75+49-12

=[-75+(-49)]+(75+49)-12

=-124+124-12

=0-12

=-12

\(-75+\left(-49\right)+|-75|+49-12=-75-49+75+49-12\)

                                                                      \(=\left[\left(75-75\right)+\left(49-49\right)\right]-12=0+0-12=-12\)

Thay A=a+b-5; B=-b-c+1 vào A+B, ta có:

A+B=(a+b-5)+(-b-c+1)

      =a+b-5-b-c+1

      =a-c-4

Thay C=b-c-4; D=b-a vào C-D, ta có:

C-D=(b-c-4)-(b-a)

      =b-c-4-b+a

      = -c+a-4

      = a-c-4

=>A+B=C-D  (đccm)

+) Ta có :

A + B = ( a + b - 5 ) +  ( -b  - c + 1 )

         = a + b - 5 - b - c + 1

         = a + ( b - b ) + c - ( 5 - 1 )

         = a + c - 4

+) C + D = ( b - c - 4 ) -  ( b - a )

             = b - c - 4 - b + a

             = ( b - b ) - c + a - 4

             = c + a - 4

Vì c + a - 4 = a + c - 4

=> A + B = C - D

15 - ( -2) - x = 7 + x

=>17 - x = 7 + x

=> x + x = 17 - 7

=> 2x = 10

=> x = 5

15-(-2)-x=7+x

15+2-x=7+x

17-x=7+x

Đổi vế

17-7=x+x

2x=10

x=10/2=5

k mình mh lại

\(17\cdot15+15\cdot65+15\cdot18\)

\(=15\left(17+65+8\right)\)

\(=15\cdot90\)

\(1350\)

\(A=\frac{3n+1}{3n-4}=\frac{3n-4+5}{3n-4}=1+\frac{5}{3n-4}\)

Suy ra : A có giá trị là số nguyên \(\Leftrightarrow\frac{5}{3n-4}\inℤ\)

\(\Leftrightarrow5⋮3n-4\left(3n-4\inℤ\right)\)

\(\Leftrightarrow3n-4\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Mà 3n - 4 chia 3 dư 2 \(\Rightarrow3n-4=-1;5\Rightarrow n=1;3\)

Vậy \(n=1;3\)

\(A=\frac{2x-1}{x+3}=\frac{2x+6-7}{x+3}=\frac{2\left(x+3\right)-7}{x+3}=2-\frac{7}{x+3}\)

Suy ra : A có giá trị nguyên \(\Leftrightarrow\frac{7}{x+3}\inℤ\)

\(\Leftrightarrow7⋮\left(x+3\inℤ\right)\)

\(\Leftrightarrow x+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow x\in\left\{-4;-2;-10;4\right\}\)

Vậy \(x=-10;-4;-2;4\)

\(A=\frac{2x-1}{x+3}=\frac{2x+6-7}{x+3}=\frac{2\left(x+3\right)-7}{x+3}=\frac{2\left(x+3\right)}{x+3}-\frac{7}{x+3}=2-\frac{7}{x+3}\)

Để \(A\inℤ\Rightarrow2-\frac{7}{x+3}\inℤ\Rightarrow\frac{7}{x+3}\inℤ\Rightarrow7⋮x+3\)

\(+,x+3=1\Rightarrow x=-2\)

\(+,x+3=-1\Rightarrow x=-4\)

\(+,x+3=7\Rightarrow x=4\)

\(+,x+3=-7\Rightarrow x=-10\)

=47+53-47-53

=0

\(\text{47-(-53)+(-47)-|-53|=47+53-47-53=(47-47)+(53-53)=0+0=0}\)