[(286:11+7.2).3-119].2002+13
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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,
\(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}=\dfrac{3}{13}\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)=\dfrac{3}{13}.0=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{36}=\left(5^3\right)^{12}=125^{12}>121^{12}=\left(11^2\right)^{12}=11^{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\)
\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)
Với \(n=k+1\)
\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)
Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)
Theo pp quy nạp ta được đpcm
\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)
Với \(n=k+1\)
\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)
Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)
Theo pp quy nạp ta được đpcm
\(1,\)
\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)
Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)
\(d,D=1^n+2^n+5^n+8^n\)
Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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}
\(\left[\left(286:11+7.2\right).3-119\right].2002+1^3\)
\(=\left[\left(26+14\right).3-119\right].2002+1\)
\(=\left[40.3-119\right].2002+1\)
\(=\left[120-119\right].2002+1\)
\(=1.2002+1\)
\(=2002+1=2003\)