Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

= \(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

= \(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

= \(\dfrac{29}{6}\)

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

= \(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

= \(\dfrac{43}{34}\)

Bài 3:

a, 800 kg 5g > 8,005 kg

b, 9 ha 4 dam² < 9\(\dfrac{5}{100}\) ha

c, 5m 5mm > 5,0005 m

d, 250 kg > \(\dfrac{1}{5}\) tấn

Bài 1

a) 3 2/5 - 1/2

= 17/5 - 1/2

= 34/10 - 5/10

= 29/10

b) 4/5 + 1/5 × 3/4

= 4/5 + 3/20

= 16/20 + 3/20

= 19/20

c) 3 1/2 × 1 1/7

= 7/2 × 8/7

= 4

d) 4 1/6 : 2 1/3

= 25/6 : 7/3

= 25/14

Bài 2

a) 3 × 1/2 + 1/4 × 1/3

= 3/2 + 1/12

= 18/12 + 1/12

= 19/12

b) 1 4/5 - 2/3 : 2 1/3

= 9/5 - 2/3 : 7/3

= 9/5 - 2/7

= 63/35 - 10/35

= 53/35

Bài 1

a) 3 2/5 - 1/2

= 17/5 - 1/2

= 34/10 - 5/10

= 29/10

b) 4/5 + 1/5 × 3/4

= 4/5 + 3/20

= 16/20 + 3/20

= 19/20

c) 3 1/2 × 1 1/7

= 7/2 × 8/7

= 4

d) 4 1/6 : 2 1/3

= 25/6 : 7/3

= 25/14

bủh

Bài 3:

a: a*S=a^2+a^3+...+a^2023

=>(a-1)*S=a^2023-a

=>\(S=\dfrac{a^{2023}-a}{a-1}\)

b: a*B=a^2-a^3+...-a^2023

=>(a+1)B=a-a^2023

=>\(B=\dfrac{a-a^{2023}}{a+1}\)

Bài 2: 

\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=9\)

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}=\dfrac{a^3+b^3}{a^3b^3}=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)}{\left(ab\right)^3}\)

\(=\dfrac{5^3-3\cdot5\cdot\left(-2\right)}{\left(-2\right)^3}=\dfrac{125+30}{8}=\dfrac{155}{8}\)

\(a-b=-\sqrt{\left(a+b\right)^2-4ab}=-\sqrt{5^2-4\cdot\left(-2\right)}=-\sqrt{33}\)

1/5

1/6

1/5

1/6

1/5

1/6

ta có: a + b=-2 ; a^2 + b^2 = 52

=> (a+b)^2 = 4 => a^2 + 2ab + b^2 = 4

=> 52 + 2ab= 4

=> 48= -2ab

=> ab= -24

a^3 + b^3 = (a+b)( a^2-ab+ b^2)

=> a^3 + b^3 = -2.(52+24)= -2. 76= -152

Ta có công thức tổng quát như sau:

\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)

Áp dụng ta có:

\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\) 

\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)

______

\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)

\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)

_____

\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)

\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)

2:

a: =4+3/8+5+2/3

=9+3/8+2/3

=216/24+9/24+16/24

=216/24+25/24

=241/24

b; =2+3/8+1+1/4+3+6/7

=6+3/8+1/4+6/7

=6+5/8+6/7

=419/56

c: \(=2+\dfrac{3}{8}-1-\dfrac{1}{4}+5+\dfrac{1}{3}\)

=6+3/8-1/4+1/3

=6+1/8+1/3

=6+11/24

=155/24

d: \(=3+\dfrac{5}{6}+6\cdot\dfrac{13}{6}\)

=3+13+5/6

=16+5/6

=101/6

e: =3+1/2+4+5/7-5-5/14

=3+4-5+1/2+5/7-5/14

=2+7/14+10/14-5/14

=2+12/14

=2+6/7=20/7

f: =9/2+1/2:11/2

=9/2+1/11

=99/22+2/22=101/22