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![](https://rs.olm.vn/images/avt/0.png?1311)
A = 5/7.(1+9/13) − 5/7.9/13
A= 5/7.(1+9/13 - 9/13)
A = 5/7.1
A = 5/7
B = 11/24 − 5/41 + 13/24 + 0.5 − 36/41
B = (11/24 + 13/24) - (5/41 + 36/41) + 0.5
B = 1 - 1 + 0.5
B = 0.5
C = −4/13.5/17 + (−12/13).4/17 + 4/13
C = 4/13.(-5/17) + (−12/13).4/17 + 4/13
C = 4/13.(-5/17 + 1) + (−12/13).4/17
C = 4/13.(−12/17) + (−12/13).4/17
C = (4.-12)/(13.17) + (−12/13).4/17
C = 4/17.(−12/13) + (−12/13).4/17
C = 4/17.(−12/13).2
C = 96/221
D = (4/3 − 3/2)2 − 2.∣−1/9∣ + (−5/18)
D = (4/3 − 3/2)2 − 2.1/9+ (−5/18)
D = -1/62 - 2/9+ (−5/18)
D = -1/12 - ( 2/9+ (−5/18) )
D = -1/12 - ( 4/18+ (−5/18) )
D = -1/12 - (-1/18)
D = -1/12 + 1/18
D = -3/36 + 2/36
D = -1/36
E = (−3/4 + 2/3):5/11 + (−1/4 + 1/3):5/11
E = (−3/4 + 2/3 + (−1/4) + 1/3):5/11
E = ((−3/4 + (−1/4)) + (2/3 + + 1/3)):5/11
E = ( - 1 + 1):5/11
E = 0:5/11
E = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)