1 và 5/27 + 7 /23 + 1/2 - 5/27 + 16/23
6 và 1/5 - ( 2/3 * 9/10 - 7/5 )
giúp mình với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 1/2 + 3/4 - (3/4 - 4 - 5)
= 1/2 + 3/4 - 3/4 + 4 + 5
= (3/4 - 3/4) + (4 + 5) + 1/2
= 0 + 9 + 1/2
= 19/2
b) [9/16 + 8/(-27)] - (19/27- 7/16 - 2)
= 9/16 - 8/27 - 19/27 + 7/16 + 2
= (9/16 + 7/16) + (-8/27 - 19/27) + 2
= 1 - 1 + 2
= 2
c) -5/8 . [4/9 + 7/(-12)]
= -5/8 . (-5/36)
= 25/288
d) 7/10 . (-3/5) + 7/10 . (-2/5) - (-3/10)
= 7/10 . (-3/5 - 2/5) + 3/10
= 7/10 . (-1) + 3/10
= -2/5
e) -3/7 . 5/9 + 4/9 . (-3/7) + 2 3/7
= -3/7 . (5/9 + 4/9) + 17/7
= -3/7 . 1 + 17/7
= 2
f) 8 2/7 - (3 4/9 + 4 2/7)
= 8 + 2/7 - 3 - 4/9 - 4 - 2/7
= (8 - 3 - 4) + (2/7 - 2/7) - 4/9
= 1 - 4/9
= 5/9
h) 3.(-1/2)² - (4/5 + 8/15) : 5/6
= 3.1/4 - 4/3 : 5/6
= 3/4 - 8/5
= -17/20
7 = 3 + 4 = √9 + √16
Do 10 > 9 nên √10 > √9
17 > 16 nên √17 > √16
⇒ √10 + √17 > √9 + √16
Vậy √10 + √17 > 7
--------
(1/8)²³ = 1/(2³)²³ = 1/2⁶⁹
(1/32)¹⁶ = 1/(2⁵)¹⁶ = 1/2⁸⁰
Do 69 < 80 nên 2⁶⁹ < 2⁸⁰
⇒ 1/2⁶⁹ > 1/2⁸⁰
Vậy (1/8)²³ > (1/³²)¹⁶
--------
5 = √25
Do 27 > 25 nên √27 > √25
Vậy √27 > 5
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
Bài 2. > , < , = ?
5/7 < 4/3 2/5 < 6/10 1/4 = 3/12 27/36 >
2/9 7/6 > 7/9
7/2 = 2/7 15/23 < 1 27/9 > 2 14/15 < 1 51/17 < 4
\(\frac{9}{8}-\frac{-11}{8}=\frac{9-\left(-11\right)}{8}=\frac{9+11}{8}=\frac{20}{8}=\frac{5}{2}\)
\(-\frac{1}{2}-\frac{5}{2}=\frac{-1-5}{2}=\frac{-6}{2}=-3\)
\(\frac{4}{3}-\frac{7}{3}=\frac{4-7}{3}=\frac{11}{3}\)
\(\frac{3}{-5}-\frac{-8}{10}=\frac{6}{-10}-\frac{-8}{10}=\frac{-6}{10}-\frac{-8}{10}=\frac{-6-\left(-8\right)}{10}=\frac{-6+8}{10}=\frac{2}{10}=\frac{1}{5}\)
\(\frac{5}{7}-\frac{-3}{21}=\frac{5}{7}-\frac{-1}{7}=\frac{5-\left(-1\right)}{7}=\frac{5+1}{7}=\frac{6}{7}\)
\(\frac{4}{-3}-\frac{9}{27}=\frac{4}{-3}-\frac{1}{3}=\frac{-4}{3}-\frac{1}{3}=\frac{-4-1}{3}=\frac{-5}{3}\)
\(\frac{7}{29}-\frac{9}{29}=\frac{7-9}{29}=\frac{2}{29}\)
\(\frac{-7}{22}-\frac{9}{22}=\frac{-7-9}{22}=\frac{-16}{22}=\frac{-8}{11}\)
\(\frac{-23}{7}-\frac{31}{7}=\frac{-23-31}{7}=\frac{-54}{7}\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
`1 5/27 + 7 /23 + 1/2 - 5/27 + 16/23`
` = 32/27 + 7/23 + 1/2 - 5/27 + 16/23 `
`= (32/27 - 5/27 ) + (7/23 + 16/23 ) + 1/2 `
`= 27/27 + 23/23 + 1/2 `
`= 1 + 1 + 1/2 `
`= 5/2`
` 6 1/5 - ( 2/3 . 9/10 - 7/5 )`
` = 31/5 - 2/3 . 9/10 + 7/5 `
`= ( 31/5 + 7/5 ) - ( 2/3 . 9/10 ) `
`= 38/5 - 3/5 `
`= 35/5`
`= 7`
`@ animephamdanhv.`