1. \(9.\left(10+20+100\right)=9.130=1170\)

2. \(14.\left(14+29+147+10000\right)=14.10190=142660\)

3. \(365.2+265.2+165.2+65.2=\left(365+265+165+65\right).2=860.2=1720\)

Bài 1:

  10 x 9 + 20 x 9 + 100 x 9

= 9 x ( 10 + 20 + 100)

= 9 x 130

= 1170

Bài 2:

14 x 14 + 14 x 29 + 14 x 147 + 14 x 10 000

= 14 x ( 14 + 29 + 147 + 10 000)

= 14 x  10190

= 142660

Bài 3:

365 + 365 + 265 + 265 + 165 + 165 + 65 + 65

= (365 + 365) + ( 265 + 265) + (165 + 165) + (65 + 65)

= 365 x 2 + 265 x 2 + 165 x 2 + 65 x 2

=  2 x ( 365 + 265 + 165 + 65)

= 2 x 860

= 1720

câu 3 có thể làm phép nhân 

Vì x=14 nên x+1=15 
Thay 15=x+1 vào A(x) Ta có:
A(x)= x^15-(x+1)x^14+(x+1)x^13-(x+1)x^12+...+(x+1)x^3-(X+1)^2+(x+1)x-15
=x^15-x^15-x^14+x^14+x^13-x^13-...+X^4+x^3-X^3-x^2+x^2-x-15
=x-15
=> A(14)=14-15=-1 
Vậy A(14)=-1 
k mình nha 

1. TÍNH NHẨM

14x10=140

1348x100=134800

5629x1000=5629000

18390 : 10=1839

328000 : 100=3280

378000 : 1000=378

14 x 10 = 140

1348 x 100 = 134800

5629 x 1000 = 5629000

18390 : 10 = 1839

328000 : 100 = 3280

378000 : 1000 = 378

Cho 1 like

144 + 12 x 13 - 15 x 12

= 12 x 12 + 12 x 13 - 15 x 12

= 12 x ( 12 + 13 - 15 )

= 12 x 10

= 120

144+12x13-15x12

= 12x12+12x13-15x12

= 12x(12+13-15)

= 12x(25-15)=12x10=120

=144+12*(15-13)

=144+12*2

=144+24

=168

1)Đặt A =  \(\frac{1}{7.8}+\frac{1}{14.10}+\frac{1}{20.13}+...+\frac{1}{38.22}\)

      \(\frac{1}{2}A=\frac{1}{8.14}+\frac{1}{14.20}+...+\frac{1}{38.44}\)

       \(\frac{1}{2}A=6\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+...+\frac{1}{38}-\frac{1}{44}\right)\)

       \(\frac{1}{2}A=6\left(\frac{1}{8}-\frac{1}{44}\right)\)

         \(\frac{1}{2}A=6.\frac{9}{88}\)

           \(\frac{1}{2}A=\frac{27}{44}\)

                \(A=\frac{1}{44}:\frac{1}{2}=\frac{1}{22}\)

2) a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)

         \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)

        \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)

       \(\frac{1}{3}-\frac{1}{9}-\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)

         \(\frac{2}{9}-\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)

                        \(\frac{1}{x\left(x+2\right)}=\frac{2}{9}-\frac{2018}{2020}\)

   Hình như đề sai . Hoặc là mình sai >:

Chọn C