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bạn phải cho ra 2 số cuối thì mới làm đc nha có 1 s
ố cuối ko làm đc đâu
A= 1-1/2 + 1-1/3 + 1/2-1/5 + 1/3-1/8+ 1/5-1/13+1/8- 1/21 +....+ 1/610- 1/1597
A= 1/610
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32 + 5. 13 - 3 . 23
= 32 + 5 .13 - 3 . 8
= 32 + 65 - 24
= 97 + 24
= 121
12.5 + 12 . 95
= 12 . ( 5 + 95 )
= 12 . 100
= 1200
b, 32+5.13-3.2^3
=32+45-3.8
=32+45+24
=97+24
=121
c,12.5+12.95
=12.(5+95)
=12.100
=1200
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\(\frac{-11^5.13^7}{11^5.13^8}=\frac{-11^4.\left(-11\right).13^7}{11^4.11.13^7.13}=\frac{-11}{11.13}=\frac{-11}{143}\)
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\(\left(-\right)\frac{1989\cdot1990+3970}{1992\cdot1991+3984}=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991+2\right)}=\frac{1989}{1993}\)
\(\left(-\right)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{3^{10}\cdot\left(-5\right)^{20}\cdot\left(-5\right)}{3^{10}\cdot\left(-5\right)^{20}\cdot3^2}=-\frac{5}{9}\)
\(\left(-\right)\frac{\left(-11\right)^5\cdot13^7}{11^5\cdot13^8}=\frac{11^5\cdot13^7\cdot\left(-1\right)}{11^5\cdot13^7\cdot13}=-\frac{1}{13}\)
\(\left(-\right)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{4}{3}\)
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\(A=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot2^3\cdot13}=\dfrac{2^{11}\cdot3\cdot13}{2^{11}\cdot13}=3\)
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\(A=\frac{4^5.13+32^2.65}{4^4.104}=\frac{2^{10}.13+2^{10}.13.5}{2^8.2^3.13}=\frac{2^{10}.13.\left(1+5\right)}{2^{11}.13}=\frac{6}{2}=3\)
Vậy A = 3
A+B