PT
1
K
Khách
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10 tháng 10 2021
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)
NV
3
26 tháng 12 2021
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
PT
0
YN
22 tháng 2 2020
\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)
Vậy A = \(\frac{-5}{7}\)
B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)
\(\Leftrightarrow B=0+\frac{-2}{11}\)
\(\Leftrightarrow\) \(B=\frac{-2}{11}\)
Vậy \(B=\frac{-2}{11}\)
@@ Học tốt
Chiyuki Fujito
K cần tk nhá
4 tháng 5 2019
\(\frac{3}{2}+\frac{-7}{9}-\frac{-1}{2}-\frac{11}{9}\)
\(=\left(\frac{3}{2}-\frac{-1}{2}\right)+\left(\frac{-7}{9}-\frac{11}{9}\right)\)
\(=\frac{3+1}{2}+\frac{-7-11}{9}\)
\(=\frac{4}{2}+\frac{-18}{9}\)
\(=2-2\)
\(=0\)
29 tháng 7 2019
a) \(A=\frac{-7}{813}+496.\left(\frac{-7}{813}\right)+\left(\frac{-7}{813}\right).316\)
\(=\frac{-7}{813}.\left(1+496+316\right)\)
\(=\frac{-7}{813}.813\)
\(=-7\)
b) \(B=\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(\frac{-9}{10}\right)\)
\(=\frac{-9}{10}.\left(\frac{5}{14}+\frac{1}{2}+\frac{1}{7}\right)\)
\(=\frac{-9}{10}.1\)
\(=\frac{-9}{10}\)
91.25 - 91.13 - 91.12 = 91.(25 - 13 - 12) = 0