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a) \(B=3+3^2+3^3+...+3^{120}\)
\(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\)
\(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\)
Suy ra B chia hết cho 3 (đpcm)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\)
\(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\)
\(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\)
Suy ra B chia hết cho 4 (đpcm)
c) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\)
\(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\)
\(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\)
Suy ra B chia hết cho 13 (đpcm)
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
\(\frac{27}{100}=0,27\) ; \(-\frac{13}{1000}=-0,013\) ; \(\frac{261}{10000}=0,0261\)
\(\frac{6}{5}=1\frac{1}{5}\) ; \(\frac{7}{3}=2\frac{1}{3}\) ; \(\frac{-16}{11}=-1\frac{5}{11}\)
a) 2,63=\(2\frac{63}{100}\)
b)\(\frac{2013}{100}\)=\(20\frac{13}{100}\)=20,13
c)\(\frac{-2}{5}=\frac{-14}{35}\)
\(\frac{-3}{7}=\frac{-15}{35}\)
vì\(\frac{-14}{35}>\frac{-15}{35}\Rightarrow\frac{-2}{5}>\frac{-3}{7}\)
\(4\frac{2}{10}m:4,2m\)
\(3\frac{12}{100}km:3,12km\)
\(2\frac{2}{25}kg:2,08kg\)
a) 5hm = 5 100 km = 0,5km
b) 12dam = 12 100 km = 0,12km
c) 64m = 64 100 km= 0,064km
d) 7dm = 7 10000 km = 0,007km