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a) [(286:11+7.2).3-199].2002+1
= [(26+14).3-199].2002+1
= [40.3-199].2002+1
= [120-199].2002+1
= -79.2002+1
= -158157
b) ko hiểu
c)1+6+11+16+...+2001+2006
số hạng của tổng là
(2006-1):2+1=1003,5
tổng
(2006+1).1003,5:2=1007012,25
d) (225-72) . (225-102) . (225-92) . (225-152)
=(152-72) . (152-102) . (152-92) . (152-152)
=176.125.144.0
0 like nha
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Dễ thế cũng hỏi ở trường tính hay lắm mà:
a) \(\dfrac{3}{7}.\dfrac{8}{11}+\dfrac{3}{7}.\dfrac{5}{11}-\dfrac{3}{7}.\dfrac{2}{11}\)
\(=\dfrac{3}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=\dfrac{3}{7}.\left(\dfrac{8+5-2}{11}\right)=\dfrac{3}{7}.\dfrac{11}{11}=\dfrac{3}{7}\)
b) \(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}\)
\(=\dfrac{3}{13}.\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)\)
\(=\dfrac{3}{13}.\left(\dfrac{8+11+6}{25}-1=\right)\dfrac{3}{13}.0=0\)
a,
\(\dfrac{3}{7}.\dfrac{8}{11}+\dfrac{3}{7}.\dfrac{5}{11}-\dfrac{3}{7}.\dfrac{2}{11}=\dfrac{3}{7}\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)=\dfrac{3}{7}.1=\dfrac{3}{7}\)
b,
\(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}=\dfrac{3}{13}\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)=\dfrac{3}{13}.0=0\)
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\(5^{36}=\left(5^3\right)^{12}=125^{12}>121^{12}=\left(11^2\right)^{12}=11^{24}\)
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Câu A là hai phân số không bằng nhau nha bạn
b: \(-\dfrac{60}{185}=\dfrac{-60:5}{185:5}=\dfrac{-12}{37}\)
c: \(\dfrac{20022002}{20032003}=\dfrac{20022002:10001}{20032003:10001}=\dfrac{2002}{2003}\)
d: \(-\dfrac{3a}{6b}=\dfrac{-a}{2b}=\dfrac{5a}{-10b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27^{11}=3^{33};81^8=3^{32}\)
\(27^{11}>81^8\)
\(5^{23}=5.5^{22}
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a: \(A=21\cdot100-11\cdot100+90\cdot100+100\cdot125\cdot16\)
\(=100\left(21-11+90\right)+100\cdot2000\)
\(=100\left(10+90+2000\right)=2100\cdot100=210000\)
b: \(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\)
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\(\left[\left(286:11+7.2\right).3-119\right].2002+1^3\)
\(=\left[\left(26+14\right).3-119\right].2002+1\)
\(=\left[40.3-119\right].2002+1\)
\(=\left[120-119\right].2002+1\)
\(=1.2002+1\)
\(=2002+1=2003\)