bài 1 Cộng các phân số sau
a ) -1/21 + -1/28
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Ta có:
\(3,1\left(21\right)=3,1+0,0\left(21\right)=\frac{31}{10}+0,0\left(01\right).21=\frac{31}{10}+\frac{0,\left(01\right)}{10}.21=\frac{31}{10}+\frac{\frac{1}{99}}{10}.21\)
\(=\frac{31}{10}+\frac{21}{9900}=\frac{3069}{9900}+\frac{21}{9900}=\frac{3090}{9900}=\frac{103}{330}\)
Bài 1:a, \(\dfrac{13}{2}\) = \(\dfrac{13\times5}{2\times5}\) = \(\dfrac{65}{10}\)
b, \(\dfrac{11}{40}\) = \(\dfrac{11\times25}{40\times25}\) = \(\dfrac{275}{1000}\)
c, \(\dfrac{21}{250}\) = \(\dfrac{21\times4}{250\times4}\) = \(\dfrac{84}{1000}\)
d, \(\dfrac{27}{45}\) = \(\dfrac{27:9}{45:9}\) = \(\dfrac{3}{5}\) = \(\dfrac{3\times2}{5\times2}\) = \(\dfrac{6}{10}\)
Bài 2:
a, (3\(\dfrac{1}{8}\) + 1\(\dfrac{3}{4}\)): 2\(\dfrac{1}{4}\)
= (\(\dfrac{25}{8}\) + \(\dfrac{7}{4}\)): \(\dfrac{9}{4}\)
= \(\dfrac{39}{8}\) \(\times\) \(\dfrac{4}{9}\)
= \(\dfrac{13}{6}\)
Bài 1:
a) \(\frac{5^2.6^{11}.16^2+6^2.11^6.15^2}{2.6^{12}.10^{14}-81^2.960^3}=\frac{5^2.2^{19}.3^{11}+2^2.3^4.11^6.5^2}{2^{27}.3^{12}.5^{14}-3^{11}.2^{18}.5^3}\)
\(=\frac{5^2.2^2.3^4.\left(2^{17}.3^7+11^6\right)}{2^{18}.3^{11}.5^3.\left(2^9.3.5^{11}-1\right)}=\frac{2^{17}.3^7+11^6}{2^{16}.3^7.5.\left(2^9.3.5^{11}-1\right)}\)
b) Đặt A = 2528 + 2524 +....+ 254 +1
=> 254.A = 2532 + 2528 +...+ 258 + 254
=> 254.A - A = 2532 -1
\(A=\frac{25^{32}-1}{25^4-1}\)
tương tự....
Thay vào:
\(\frac{25^{28}+25^{24}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}=\frac{\frac{25^{32}-1}{25^4-1}}{\frac{25^{32}-1}{25^2-1}}=\frac{25^2-1}{25^4-1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
Bài 3:
a: =>x/56=3/7
hay x=21
b: =>3/x=6/14
hay x=7
c: =>35/x=7/15
hay x=75
d: =>2x=-10/9
hay x=-10/18=-5/9
ko bit loi giai
\(-\frac{1}{21}+\left(-\frac{1}{28}\right)\)
\(=-\frac{1}{21}-\frac{1}{28}\)
\(=-\frac{1}{12}\)
Vậy -1/21 + -1/28 = -1/12
ủng hộ nha