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Đặt A=2/3+2/6+2/12+...+2/768
=2/3(1+1/2+1/4+...+1/256)
Đặt B=1+1/2+1/4+...+1/256
=>2B=2+1+1/2+...+1/128
=>B=2-1/256=511/256
=>\(A=\dfrac{2}{3}\cdot\dfrac{511}{256}=\dfrac{511}{128\cdot3}=\dfrac{511}{384}\)
60 2 84 2 215 3 400 2
30 2 42 2 75 3 200 2
15 3 21 3 25 5 100 2
5 5 7 7 5 5 50 2
1 1 1 25 5
5 5
1
máy không ghi dấu gạch được nên thông cảm
học tốt nhé!
Lời giải:
$b=2021^0=1< 26^9=a$
$c=26^9=(2.13)^9=2^9.13^9=(2^3)^3.13^9=8^3.13^9<13^3.13^9$
$=13^{12}< 13^{15}=a$
Vậy $b< c< a$
Đáp án C.
Hỗn số (3 1/3 + 2,5) : ( hỗn số 3 1/6 - hỗn số 4 1/5 ) - 11/31
=(10/3+2,5)/(19/6-21/5)-11/31
=(17,5/3)/(-31/30)-11/31
=-175/31-11/31
=-186/31
=-6
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
`4 / 7 xx [-5] / 6 xx 7 / 4 xx 12 / 5`
`= [ 4 xx (-5) xx 7 xx 12 ] / [ 7 xx 6 xx 4 xx 5 ]`
`= [ 4 xx (-5) xx 7 xx 6 xx 2 ] / [ 7 xx 6 xx 4 xx 5 ]`
`= -2`
4/7 × -5/7 × 7/4 ×12/5
=> -10/21 × 42/10
=> -420/210
=> -42/21
Đáp số: -42/21
p: \(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+...+\dfrac{5}{50\cdot51}\)
\(=5\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{50\cdot51}\right)\)
\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(=5\cdot\left(1-\dfrac{1}{51}\right)=5\cdot\dfrac{50}{51}=\dfrac{250}{51}\)
q: \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{210}\)
\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{420}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\right)\)
\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{20\cdot21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{42}=\dfrac{19}{21}\)
[31.(x+3)].12=144
31.(x+3) =144:12
31.(x+3) = 12
x+3 = 12:31
x+3 = 12/31
x = 12/31 - 3
x = -81/31
[31 . (x + 3)] . 12 = 144
31 . (x + 3) = 144 : 12
31 . (x + 3) = 12
x + 3 = \(\frac{12}{31}\)
x = \(\frac{12}{31}-3=\frac{12}{31}-\frac{93}{31}\)
x = \(\frac{-81}{31}\)
Vậy x = \(\frac{-81}{31}\)