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a, \(M=1+6+6^2+6^3+...+6^{99}\)
\(M=6\cdot(1+6)+6^2(1+6)+6^3(1+6)+...+6^{99}(1+6)\)
\(M=6\cdot7+6^2\cdot7+6^3\cdot7+...+6^{99}\cdot7\)
\(M=7\cdot\left[6+6^2+6^3+...+6^{99}\right]⋮7(đpcm)\)
b, \(M=1+6+6^2+6^3+...+6^{99}\)
\(M=6\cdot\left[1+6+6^2+6^3\right]+...+6^{96}\left[1+6+6^2+6^3\right]\)
\(M=6\cdot\left[7+36+216\right]+...+6^{96}\left[7+36+216\right]\)
\(M=6\cdot259+...+6^{96}\cdot259\)
\(M=259\cdot\left[6+...+6^{96}\right]⋮259\)
Vậy \(M⋮259(đpcm)\)
\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+2^{20}.3^{20}}{2^{20}.3^{20}-3^{20}+3^{40}}\)
\(=\frac{2^{20}\left(2^{20}-1+3^{20}\right)}{3^{20}\left(2^{20}-1+3^{20}\right)}\)
\(=\frac{2^{20}}{3^{20}}\)
Mình nghĩ sửa 3 thành 1 sẽ hợp lí hơn
a)\(S=1+3^2+3^4+...+3^{2002}\)
=>\(3^2.S=3^2+3^4+3^6+...+3^{2004}\)
=>\(9S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(1+3^2+3^4+...+3^{2002}\right)\)
=>\(8S=3^{2004}-1\)
=>\(S=\frac{3^{2004}-1}{8}\)
b)\(S=1+3^2+3^4+...+3^{2002}\)
=>\(S=\left(1+3^2+3^4\right)+...+\left(3^{1998}+3^{2000}+3^{2002}\right)\)
=>\(S=91+...+3^{1998}\left(1+3^2+3^4\right)\)
=>\(S=91+...+3^{1998}.91\)
=>\(S=91\left(1+...+3^{1998}\right)\)
=>\(S=7.13.\left(1+...+3^{1998}\right)\) chia hết cho 7 (đpcm)
a) 8x+2x = 25.22
8x+2x = 25. 4
8x+2x = 100
( 8 + 2) . x = 100
10 . x =100
x =100 :10
x =10
1) 6x+x=28 => 7x=28 => x= 28:7=4.
2) \(5x+x=39-3^{11}:3^9=>6x=39-3^2=>6x=39-9=>6x=30=>x=30:6=5\)
3) \(7x-x=5^{21}:5^{19}+3.2^2-7^0.7\)
\(6x=5^2+3.4-1.7=>6x=25+12-7=>6x=30=>x=30:6=5.\)
4)\(7x-2x=6^{17}:6^{15}+44:11=>5x=6^2+4=>5x=36+4=>5x=40=>x=40:5=8.\)
7x-x=521:519+3.22-70
=> 6x = 5^2 + 12 -1
=> 6x = 36
=> x = 36/6 = 6
Kết quả 6
Học tốt
Đáp án:
B=62021−65B=62021−65
Giải thích các bước giải:
B=61+62+63+⋯+62020B=61+62+63+⋯+62020
→6B=62+63+64+⋯+62021→6B=62+63+64+⋯+62021
→6B−B=(62+63+64+⋯+62021)−(61+62+63+⋯+62020)→6B−B=(62+63+64+⋯+62021)−(61+62+63+⋯+62020)
→5B=62021−6→5B=62021−6
→B=62021−65
mình chưa hiểu rõ