a)1001^2=(1000+1)^2=1000^2+2000+1^2=1000000+2001=1002001

b)=(30-0.1)*(30+0.1)=30^2-1^2=900-1=899

c)=(200-1)^2=200^2-400+1^2=40000-401=39599

d)=(84-16)*(84+16)=68*100=6800

e)=(313-312)*(313+312)=1*625=625

f)=(50-3)*(50+3)=50^2-3^2=550-9=541

Chúc bạn học tốt!

a) \(1001^2=\left(1000+1\right)^2=1000^2+2.1000.1+1^2=1002001\)

b) \(29,9\times30,1=\left(30-0,1\right).\left(30+0,1\right)=30^2-\left(0,1\right)^2=899,99\)

c) \(\left(31,8\right)^2-2.31,8.21,8+\left(21,8\right)^2=\left(31,8-21,8\right)^2=10^2=100\)

a)1002001
b)899,99
c)100
d)-43

a) 1001²  = 1002001
b) 29,9 . 30,1 = 899,99
c) (31,8 )² - 2 . 31,8 . 21,8 + (21,8 )² = 100

18?

190:))

190?

0.13427871148

\(\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{1}{7}\right)=\dfrac{1}{3}\times1=\dfrac{1}{3}\)

1/3

Ta có: \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)

\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

Giải:

1/12+1/20+1/30+...+1/9702

=1/3.4+1/4.5+1/5.6+...+1/98.99

=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99

=1/3-1/99

=32/99

Chúc bạn học tốt!

Ta sử dụng t/c sau:

`1/(a(a+1))=1/a-1/(a+1)`

`=>1/12+1/20+1/30+...+1/9072`

`=1/(3.4)+1/(4.5)+....+1/(98.99)`

`=1/3-1/4+1/4-1/5+....+1/98-1/99`

`=1/3-1/99`

`=32/99`

=1/6-1/7+1/7-1/8+1/8-1/9+...+1/10-1/11

=1/6-1/11=5/66