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 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 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 1: Tính
a, 1/2 x 2/3 x 3/4 = 1/4
b, 7/8 x 8/9 x 9/10 = 7/10
c, 5/14 x 7/15 x 28/7 = 2/3
d, 2 x 1/2 x 3 x 1/3 x 4 x 1/4 x 5 x 1/5 = 1
Bài 2: Tính
a, 7/20 - ( 5/8 - 3/5 ) = 7/20 - 1/40 = 14/40 - 1/40 = 13/40
b, 5/6 + ( 5/9 - 1/4 ) = 5/6 + 11/36 = 30/36 + 11/36 = 41/36
c, 9/10 - ( 2/5 + 3/10 ) + 7/20 = 9/10 - 7/10 + 7/20 = 2/10 + 7/20 = 4/20 + 7/20 = 11/20.
Bài 3: Tìm x:
a, 1/2 + x = 5/6
x = 5/6 - 1/2
x = 1/3
b, 5/6 - x = 1/3
x = 5/6 - 1/3
x = 1/2
c, x - 1/mấy vậy bạn
d, 1/3 x x = 1/6
x = 1/6 : 1/3
x = 1/2

c/
C = 1/100-1/100-1/99-1/99-1/98-1/98-1/97-..........-1/3-1/2-1/2-1/1
C = 1/100-1/100-1/1
C = 0-1/1
C = -1

5A=5+25+5^4+.....+5^100+5^101

5A-A=5^101-1

=>A=5^101-1

A=1+5+5^2+5^3+...+5^99+5^100

\(5A=5+5^2+5^3+...+5^{100}+5^{101}\)

\(5A-A=5^{101}-1\)

\(4A=5^{101}-1\)

\(A=\frac{5^{101}-1}{4}\) 

a. 

\(\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{12}\right)+\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{12}\right)\) 

\(=\left(\frac{5}{4}+\frac{3}{5}+\frac{3}{8}-\frac{7}{5}\right):\left(-\frac{11}{12}\right)\)  

\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{12}\right)\) 

\(=\frac{33}{40}:\left(-\frac{11}{12}\right)\) 

\(=\frac{33}{40}\cdot\left(-\frac{12}{11}\right)\) 

\(=\frac{-9}{10}\)  

b. 

\(\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{15}\right)+\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\) 

\(=\left(\frac{3}{8}-\frac{7}{5}+\frac{5}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)  

\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{15}\right)\) 

\(=\frac{33}{40}:\left(-\frac{11}{15}\right)\) 

\(=\frac{33}{40}\cdot\left(-\frac{15}{11}\right)\) 

\(=\frac{-9}{8}\)

bạn nào trảlời nhanh mình k cho

mình cảm ơn trc 

Ta có: \(a+\frac{1}{a}=3\Leftrightarrow a^2+\frac{1}{a^2}+2=9\Leftrightarrow a^2+\frac{1}{a^2}=7\Leftrightarrow a^4+\frac{1}{a^4}+2=49\Leftrightarrow a^4+\frac{1}{a^4}=47\)

Lại có: \(\left(a+\frac{1}{a}\right)\left(a^4+\frac{1}{a^4}\right)=3.47\Leftrightarrow a^5+\frac{1}{a^5}+a^3+\frac{1}{a^3}=141\left(1\right)\)

Mặt khác \(\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right)=3.7\Leftrightarrow a^3+\frac{1}{a^3}+a+\frac{1}{a}=21\Leftrightarrow a^3+\frac{1}{a^3}=18\)(2)

Thay (2) vào (1) ta đc:

\(a^5+\frac{1}{a^5}+18=141\Leftrightarrow a^5+\frac{1}{a^5}=123\)

Câu 1: Kéo dài Kc cắt AD tại N 
Ta có AB//CD nên góc BAD = góc CDN =60o60o
K đối xứng vs F qua BC nên góc BCD = góc KCB = 60o60o
\Rightarrow góc CND = góc KCB = 60 độ 
\Rightarrow tam giác CND đều \Rightarrow CN= DN
Lại có CK= DE ( cùng = CF)
\Rightarrow KN=EN \Rightarrow tam giác KNE đều 
\Rightarrow góc KEN= góc CDN = 60 độ
\Rightarrow KE//CD//AB