Ta có: \(\frac{x+1}{15}+\frac{x+2}{16}+\frac{x+3}{17}=3\)

=> \(\left(\frac{x+1}{15}-1\right)+\left(\frac{x+2}{16}-1\right)+\left(\frac{x+3}{17}-1\right)=0\)

=> \(\frac{x-14}{15}+\frac{x-14}{16}+\frac{x-14}{17}=0\)

=> \(\left(x-14\right)\left(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}\right)=0\)

=> \(x-14=0\)

=> \(x=14\)

3x+751/2040=0

x=-751/6120

\(\frac{x+1}{18}+\frac{x+2}{17}=\frac{x+3}{16}+\frac{x+4}{15}\)

\(\Rightarrow\frac{x+1}{18}+1+\frac{x+2}{17}+1=\frac{x+3}{16}+1+\frac{x+4}{15}+1\)

\(\Rightarrow\frac{x+19}{18}+\frac{x+19}{17}-\frac{x+19}{16}-\frac{x+19}{15}=0\)

\(\Rightarrow\left(x+19\right).\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)

Vì \(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\ne0\)

Nên \(x+19=0\)

\(\Rightarrow x=-19\)

Vậy x = -19

b)Tận cùng=5 hoặc 0 nhưng mình ngại viết lắm,thông cảm nha

Tính ngoặc tròn 

\(4\cdot8-16\cdot2\) 

\(=32-32\) 

\(=0\)                                    

Vậy tích trên bằng 0 ( vì có 1 thừa số = 0 ) 

                                                 Bài giải

1 x 2 x 3 x 4 x 5 x 6 x 7 x ( 4 x 8 - 16 x 2 ) x 15 x 16 x 17 ... x 2019 x 2020 

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x ( 32 - 32 ) x 15 x 16 x 17 ... x 2019 x 2020 

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 0 x 15 x 16 x 17 ... x 2019 x 2020 

= 0

chắt do tui quên tính 4 x 8 - 16 x 2 nên mới vậy :v

Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16

7+x=-16

    x=-16-7

    x=-23
2) 2x – 35 = 15

2x=15+35

2x=50

  x=50:2

  x=25
3) 3x + 17 = 12

3x=12-17

3x=-5

  x=-5/3
4) (2x – 5) + 17 = 6

2x-5=6-17

2x-5=-11

2x=-11+5

2x=-6

  x=-6:2

  x=-3
5) 10 – 2(4 – 3x) = -4

2(4-3x)=10-(-4)

2(4-3x)=14

4-3x=14:2

4-3x=7

3x=4-7

3x=-3

  x=-3:3

  x=-1
6) - 12 + 3(-x + 7) = -18

3(-x+7)=-18-(-12)

3(x+7)=-6

x+7=-6:3

x+7=-2

    x=-2-7

    x=-9

tự đi mà làm

2, 

a) \(315-\left(135-x\right)=215\)

\(\Rightarrow135-x=315-215\)

\(\Rightarrow135-x=100\)

\(\Rightarrow x=135-100\)

\(\Rightarrow x=35\)

b) \(x-320:32=25\cdot16\)

\(\Rightarrow x-10=5^2\cdot4^2\)

\(\Rightarrow x-10=20^2\)

\(\Rightarrow x-10=400\)

\(\Rightarrow x=410\)

c) \(3\cdot x-2018:2=23\)

\(=3\cdot x-1009=23\)

\(\Rightarrow3\cdot x=1032\)

\(\Rightarrow x=1032:3\)

\(\Rightarrow x=344\)

d) \(280-9\cdot x-x=80\)

\(\Rightarrow280-x\cdot\left(9+1\right)=80\)

\(\Rightarrow280-10\cdot x=80\)

\(\Rightarrow10\cdot x=280-80\)

\(\Rightarrow10\cdot x=200\)

\(\Rightarrow x=20\)

e) \(38\cdot x-12\cdot x-x\cdot16=40\)

\(\Rightarrow x\cdot\left(38-12-16\right)=40\)

\(\Rightarrow x\cdot10=40\)

\(\Rightarrow x=40:10\)

\(\Rightarrow x=4\)

bài 1 cậu ko giải giúp mình dc à