A) He gets up at seven o'clock in the morning.
B) Bob often washes the car.
C) Jane was born in 1964.
D) We seldom have lunch at noon.
E) What time do you leave for work?
F) John never goes to bed early.
G) How often do you clean your room?
H) I sometimes go out with my friends on Saturdays.
I) He goes to the swimming pool twice a week.
J) We always go for a walk on Sunday afternoons.
k mik nha!Thanks

a ) He gets up at seven o'clock in the morning

b)Bob often washes the car

c)jane was born in 1964

d)we seldom have lunch at noon

e

Thiếu chủ ngữ

a.thiếu chủ ngữ

Bạn kiểm tra lại đề bài giúp mình nhé!

đề bài đúng rồi bạn ạ

4

a)\(S=1+2+2^2+...+2^{10}\)

\(2S=2+2^2+2^3+...+2^{11}\)

\(2S-S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)

\(S=2^{11}-1\)

b)\(S=1+3+3^2+...+3^6\)

\(3S=3+3^2+3^3+...+3^7\)

\(3S-S=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)

\(2S=3^7-1\)

\(S=\frac{3^7-1}{2}\)

a.\(S=1+2+2^2+...+2^{10}\)

\(2S=2+2^2+2^3+...+2^{11}\)

\(\Rightarrow2S-S=S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)

\(=2^{11}-1\)

b) \(S=1+3+3^2+...+3^6\)

\(3S=3+3^2+3^3+...+3^7\)

\(\Rightarrow3S-S=2S=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)

\(2S=3^7-1\Rightarrow S=\frac{3^7-1}{2}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)

\(1-\frac{1}{x+1}=\frac{4}{5}\)

\(\frac{x}{x+1}=\frac{4}{5}\)

\(\frac{x}{x+1}=\frac{4}{4+1}\)

\(\Rightarrow x=4\)

Vậy x = 4

=))

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{4}{5}\)

\(\Leftrightarrow\frac{1}{x-1}=1-\frac{4}{5}\)

\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{5}\)

\(\Leftrightarrow x-1=5\)

\(\Leftrightarrow x=5+1\)

\(\Leftrightarrow x=6\)

~ Rất vui vì giúp đc bn ~ ^_<

Tổng của ba số t1; t2; t3 là:

(427+688+517):2=816

Số t3 là:

816-427=389

Số t1 là:

816-688=128

Số t2 là:

816-517=299

\(\left(1+2\frac{1}{4}-3\frac{1}{3}\right):\left(1+3\frac{1}{2}-4\frac{7}{12}\right)\)

\(=\left(1+2-3+\frac{1}{4}-\frac{1}{3}\right):\left(1+3-4+\frac{1}{2}-\frac{7}{12}\right)\)

\(=\frac{-1}{12}:\frac{-1}{12}=\frac{-1}{12}.\frac{12}{-1}=\frac{12}{12}=1\)

a. (1 + \(2\frac{1}{4}\) -  \(3\frac{1}{3}\)) : (1 + \(3\frac{1}{2}\) - \(4\frac{7}{12}\))

= \(-\frac{1}{12}\): ( \(-\frac{1}{12}\))

= 1

b. \(5\frac{1}{2}\)-  \(14\frac{3}{7}\) :  \(\frac{9}{13}\)-  \(3\frac{4}{7}\) :  \(\frac{9}{13}\)

= \(\frac{11}{2}\)+ (\(14\frac{3}{7}-3\frac{4}{7}\)) : \(\frac{9}{13}\)

= \(\frac{11}{2}\)- 26

= \(-\frac{41}{2}\)

#)Giải :

Ta có : \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2007.2008}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2007}-\frac{1}{2008}=\frac{1}{5}-\frac{1}{2008}=\frac{2003}{10004}>\frac{1}{5}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2007^2}>\frac{1}{5}\)

\(\frac{1}{5}-\frac{1}{2018}>\frac{1}{5}????\)