\(A=\left[2\cdot3^{15}\cdot3^8-5\cdot3^2\cdot3^{10}\right]\cdot\dfrac{998}{1993006}-1817\)

\(=\left[3^{23}\cdot2-5\cdot3^{12}\right]\cdot\dfrac{998}{1993006}-1817\)

\(=3^{12}\cdot\left[3^{11}\cdot2-5\right]\cdot\dfrac{998}{1993006}-1817\)

\(=\dfrac{1}{1997}\cdot3^{12}\cdot354289-1817\)

\(\simeq94281458.14\)

a) A=1-2-3+4+5-6-7+.....+1996+1997-1998-1999+2000

=(1-2-3+4)+(5-6-7+8)+...+(1997-1998-1999+2000)

=0

b) B=1-3+5-7+....+2001-2003+2005

=(1-3)+(5-7)+...+(2001-2003)+2005

=-2.501+2005

=-1002+2005

=1003

c) C=1-2-3+4+5-6-7+8+.....+1993-1994-1995+1996+1997

=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+1997

=1997

d) D=1000+998+996+......+10-999-997-995-...-11

=(1000-999)+(998-997)+(996-995)+....+(12-11)+10

=1.495+10

=595

Bài toán này giống của lớp 7 ghê

lớp 6 đó

1001

Yêu cầu bài toán chỉ đơn thuần tính cái này thôi à em!

\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+...+\dfrac{1}{999}.\dfrac{1}{1000}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\\ =1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

ta có

1/1.1/2=1-1/2

1/2.1/3=1/2-1/3

1/3.1/4=1/3-1/4

............

1/999.1/1000=1/999-1/1000

Từ đó suy ra

1/1.1/2+1/2-1/3+1/3+.......+1/998.1/999+1/999.1/1000

=1/1-1/2+1/2-1/3+1/3-.....+1/998-1/999+1/999-1/1000

=1-1/1000

=1000/1000-1/1000

=999/1000

nhớ like bạn nhé

 \(2^3. 4^2 . 5^3 \)\(= 8 . 16 . 125 = 16000\)

\(9.3^3.\dfrac{1}{81}.3^3=\left(9.81\right)\left(\dfrac{1}{81}.81\right)=729\)

\(3^2.5^2\left(\dfrac{2}{3}\right)^2=9.25.\dfrac{2^2}{3^2}=9.25.\dfrac{4}{9}=100\)

\(\left(\dfrac{1}{3}\right)^2.\dfrac{1}{3}.9^2=\dfrac{1}{9}.9.\dfrac{1}{3}.9=3\)

\(2^3.4^2.5^3=\left(2^3.5^3\right).4^2=10^3.4^2=1000.16=16000\)

\(9.3^3.\dfrac{1}{81}.3^2=\left(9.\dfrac{1}{81}\right).\left(3^3.3^2\right)=\dfrac{1}{9}.3^5=\dfrac{3^5}{3^2}=3^3=81\)

\(3^2.5^2.\left(\dfrac{2}{3}\right)^2=5^2.\left[3^2.\dfrac{2^2}{3^2}\right]=5^2.2^2=10^2=100\)

\(\left(\dfrac{1}{3}\right)^2.\dfrac{1}{3}.9^2=\left(\dfrac{1}{3}\right)^2.\left(\dfrac{1}{3}.9^2\right)=\dfrac{1}{9}.\dfrac{9^2}{3}=\dfrac{9}{3}=3\)

a) Ta có: \(2^3\cdot4^2\cdot5^3\)

\(=2^3\cdot2^4\cdot5^3\)

\(=2^5\cdot5^3\)

\(=8\cdot125=1000\)

b) Ta có: \(9\cdot3^3\cdot\dfrac{1}{81}\cdot3^2\)

\(=3^5\cdot3^2\cdot\dfrac{1}{3^4}\)

\(=3^7\cdot\dfrac{1}{3^4}\)

\(=\dfrac{3^7}{3^4}=3^3=27\)

c) Ta có: \(3^2\cdot5^2\cdot\left(\dfrac{2}{3}\right)^2\)

\(=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2\)

\(=\left(5\cdot2\right)^2=10^2=100\)

d) Ta có: \(\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3}\cdot9^2\)

\(=\dfrac{1}{9}\cdot\dfrac{1}{3}\cdot9^2\)

\(=\dfrac{1}{27}\cdot81=3\)