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b) -5/7 . 4/13 + -5/7 . 9/13 + -2/7
= -5/7 . (4/13 + 9/13) + -2/7
= - 5/7 + - 2/7
= -1
k nếu đúng, học tốt nha
Bài làm :
a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)
b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)
c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)
d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)
e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)
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
Ta có \(A=\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{102}{5^{102}}+\frac{103}{5^{103}}\)
=> 5A = \(\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{102}{5^{101}}+\frac{103}{5^{102}}\)
Khi đó 5A - A = \(\left(\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{102}{5^{101}}+\frac{103}{5^{102}}\right)-\left(\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{102}{5^{102}}+\frac{103}{5^{103}}\right)\)
=> 4A = \(\frac{3}{5^2}+\left(\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{102}}\right)-\frac{103}{5^{103}}\)
=> 4A = \(\frac{3}{5^2}+\frac{\frac{1}{5^2}-\frac{1}{5^{102}}}{4}-\frac{103}{5^{103}}\)
=> A = \(\frac{3}{5^2.4}+\left(\frac{1}{5^2}-\frac{1}{5^{102}}\right).\frac{1}{16}-\frac{103}{5^{103}.4}\)
=> A = \(\frac{3}{100}+\frac{1}{5^2}.\frac{1}{16}\left(1-\frac{1}{5^{100}}\right)-\frac{103}{5^{103}.4}=\frac{3}{100}+\frac{1}{400}\left(1-\frac{1}{5^{100}}\right)-\frac{103}{5^{103}.4}\)
\(=\frac{3}{100}+\frac{1}{400}-\frac{1}{400.5^{100}}-\frac{103}{5^{103}.4}=\frac{13}{400}-\frac{1}{400.5^{100}}-\frac{103}{5^{103}.4}< \frac{13}{400}\left(\text{ĐPCM}\right)\)
Vậy \(A< \frac{13}{400}\)(đpcm)