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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)
b, Ta có: 125^80 = (5^2)^80 =5^160
25^118=(5^2)^118=5^236
vì 5^160<5^236 nên :125^80<5^118
( kí hiệu này ^ là mũ )
a)\(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
\(=\dfrac{-5}{9}+\dfrac{5}{9}:\left(\dfrac{20}{12}-\dfrac{25}{12}\right)\)
\(=\dfrac{-5}{9}+\dfrac{5}{9}\cdot\dfrac{-12}{5}\)
\(=\dfrac{-5}{9}+\dfrac{-12}{9}\)
\(=\dfrac{-17}{9}\)
b) \(\dfrac{-7}{25}\cdot\dfrac{11}{13}+\dfrac{-7}{25}\cdot\dfrac{2}{13}-\dfrac{18}{25}\)
\(=\dfrac{-7}{25}\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
\(=\dfrac{-7}{25}\cdot1-\dfrac{18}{25}\)
\(=\dfrac{-25}{25}\)
\(=-1\)
c) \(\left(\dfrac{5}{7}\cdot0,6-5:3\dfrac{1}{2}\right)\cdot\left(40\%-1,4\right)\cdot\left(-2\right)^3\)
\(=\left(\dfrac{3}{7}-\dfrac{10}{7}\right)\cdot\left(0,4-1,4\right)\cdot\left(-8\right)\)
\(=\left(-1\right)\cdot\left(-1\right)\cdot\left(-8\right)\)
\(=-8\)
a, \(\left(2\dfrac{3}{5}-3\dfrac{5}{9}\right):\left(3\dfrac{10}{21}-1\dfrac{3}{7}\right)\)
\(=\dfrac{-43}{45}:\dfrac{43}{21}=\dfrac{-43}{45}.\dfrac{21}{43}=\dfrac{-7}{15}\)
b, \(5\dfrac{1}{2}-14\dfrac{3}{7}:\dfrac{9}{13}-3\dfrac{4}{7}:\dfrac{9}{13}\)
\(=5\dfrac{1}{2}-14\dfrac{3}{7}.\dfrac{13}{9}-3\dfrac{4}{7}.\dfrac{13}{9}\)
\(=5\dfrac{1}{2}-\dfrac{13}{9}.\left(14\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=5\dfrac{1}{2}-\dfrac{13}{9}.15=5\dfrac{1}{2}-\dfrac{65}{3}\)
\(=\dfrac{-97}{6}\)
Chúc bạn học tốt!!!
CÁCH 1 : A = \(\dfrac{235}{11}-\left(\dfrac{8}{5}+\dfrac{81}{11}\right)\)
A = \(\dfrac{235}{11}-\left(\dfrac{88}{55}+\dfrac{405}{55}\right)\)
A = \(\dfrac{235}{11}-\dfrac{493}{55}\)
A = \(\dfrac{1175}{55}+\dfrac{493}{55}\)
A = \(\dfrac{1668}{55}\)
a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)
b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)
d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)
e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)
\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)
f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)
g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)
h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)
i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)
Thực hiện phép tính ta được A = l0 ; B = 6.
Từ đó suy ra A > B.