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`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`-7/25 + (-8)/25`
`= (-7 - 8)/25`
`= -15/25`
`= -3/5`
`b)`
`6/13 + (-15)/39`
`= 18/39 + (-15)/39`
`= (18 - 15)/39`
`= 3/39`
`= 1/13`
`c)`
`5/7 + 4/(-14)`
`= 10/14 + (-4)/14`
`= (10 - 4)/14`
`= 6/14`
`= 3/7`
`d)`
`-8/18 + (-15)/27`
`= -4/9 + (-5)/9`
`= (-4-5)/9`
`= -9/9 = -1`
`2,`
`a)`
`3/5 + (-7)/4`
`= 12/20 + (-35)/20`
`= (12 - 35)/20`
`=-23/20`
`b)`
`(-2) + (-5)/8`
`= (-16)/8 + (-5)/8`
`= (-16 - 5)/8`
`= -21/8`
`c)`
`1/8 + (-5)/9`
`= 9/72 + (-40)/72`
`= (9-40)/72`
`= -31/72`
`d)`
`6/13 + (-14)/39`
`= 18/39 + (-14)/39`
`= (18 - 14)/39`
`= 4/39`
`e)`
`(-18)/24 + 15/21`
`= (-3)/4 + 5/7`
`= (-21)/28 + 20/28`
`= (-21 + 20)/28`
`= -1/28`
a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2
b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2
c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)
d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45
e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)
f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) . ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)
g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)
h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)
i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\)
= \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\) = \(\dfrac{-193}{200}\)
j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )
= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\)
= -1+\(\dfrac{-13}{21}\)
= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)
Khôi nguyễn
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
a)( \(\dfrac{-8}{40}\)+\(\dfrac{25}{40}\))+(\(\dfrac{-3}{8}\))
=\(\dfrac{17}{40}\)+(\(\dfrac{-3}{8}\))
=\(\dfrac{17}{40}\)+(\(\dfrac{-15}{40}\))
=\(\dfrac{2}{40}\)
=\(\dfrac{1}{20}\)
b)\(\dfrac{-5}{21}\)-(\(\dfrac{16}{21}\)-\(\dfrac{21}{21}\))
=\(\dfrac{-5}{21}\)-\(-5\)
=\(\dfrac{-25}{105}\)-\(\dfrac{-105}{105}\)
=\(\dfrac{16}{21}\)
bằng đây thôi nha mỏi tay quá
c: =3/7(-5/11+6/11)=3/7*1/11=3/77
d:=8/-3(4/3-7/3)=8/3
e: =3,4-5/8-0,4-3/8
=2-1
=1
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
A = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +....+ 99 - 100
A = (1 - 2) + ( 3- 4) + ....+ (99 - 100)
Xét dãy số 1; 3;...; 99
Dãy số trên là dãy số cách đều với khoảng cách là: 3 - 1 = 2
Số số hạng của dãy số trên là: ( 99 - 1): 2 + 1 = 50
A là tổng của 50 nhóm mỗi nhóm cóa giá tri là: 1 - 2 = - 1
A = - 1 \(\times\) 50 = - 50
B = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 +...+ 97 - 98 - 99 + 100
B = ( 1 - 2 - 3 + 4) + ( 5 - 6 - 7 + 8) +...+ ( 97 - 98 - 99 + 100)
B = 0 + 0 +...+ 0
B = 0