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Thực hiện phép tính ( bằng cách hợp lí nếu có thể):
a, \(5\frac{4}{13}.15\frac{3}{41}-5\frac{4}{13}.2\frac{3}{41}\)
\(=5\frac{4}{13}\left(15\frac{3}{41}-2\frac{3}{41}\right)\)
\(=15\frac{4}{13}\left(\frac{618}{41}-\frac{85}{41}\right)\)
\(=\frac{69}{13}.13\)
\(=69\)
b, \(6.\left(-\frac{1}{3}\right)^2-\left(\frac{1}{4}:2-\frac{7}{16}.\frac{-4}{21}\right)\)
\(=6.\frac{1}{9}-\left(\frac{1}{8}-\frac{-1}{12}\right)\)
\(=\frac{2}{3}-\left(\frac{3}{24}-\frac{-2}{24}\right)\)
\(=\frac{2}{3}-\frac{5}{24}\)
\(=\frac{16}{24}-\frac{5}{24}\)
\(=\frac{11}{24}\)
Chúc bạn hok tốt!!! lưu khánh huyền

a) -90/189 + 45/84 - 78/126
= -10/21 + 15/28 - 13/21
= (-10/21 - 13/21) + 15/28
= -24/21 + 15/28
= -17/28

\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)
\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)
\(A=1-1=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)
\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)
\(A=0-\frac{43}{101}=\frac{-43}{101}\)
\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)
\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)
\(C=\frac{-2}{1}=-2\)

\(2^{41}+4^{21}+8^{15}=2^{41}+\left(2^2\right)^{21}+\left(2^3\right)^{15}=2^{41}+2^{42}+2^{45}=2^{41}\left(1+2+2^4\right)=2^{41}.19\) chia hết cho 19(đpcm)

\(\left(1+11+21+31+41+27+37+47\right)^{11}-\left(45+27+17\right)^6⋮2\)
mk ghi lại đề nha

\(\text{}\text{}\)\(=\frac{27}{43}.\frac{34}{58}-\frac{21}{41}.\frac{1}{2}+\frac{9}{58}:\frac{43}{37}-\frac{6}{29}:\frac{41}{21}\\ =\frac{27}{43}.\frac{34}{58}-\frac{21}{41}.\frac{1}{2}+\frac{9}{58}.\frac{37}{43}-\frac{6}{29}.\frac{21}{41}\)

a) \(21^{10}-1=\left(21^5\right)^2-1^2=\left(21^5+1\right).\left(21^5-1\right)\)
\(21^5+1=\overline{...1}=2k+1+1=2n\)
\(21^5-1=\overline{...01}-1=\overline{...00}\)
\(\Rightarrow21^{10}-1=2n.\overline{...00}⋮200\left(đpcm\right).\)
b) \(39\equiv-1\left(mod40\right)\)
\(\Rightarrow39^{20}\equiv1\left(mod40\right)\)
\(\Rightarrow39^{19}\equiv-1\left(mod40\right)\)
\(\Rightarrow39^{20}+39^{19}\equiv1+\left(-1\right)\left(mod40\right)\)
\(\Leftrightarrow39^{20}+39^{19}\equiv0\left(mod40\right)\)
\(\Rightarrow39^{20}+39^{19}⋮40\left(đpcm\right).\)
d) \(2005\equiv-1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}\equiv\left(-1\right)^{2007}=-1\left(mod2006\right)\)
\(2007\equiv1\left(mod2006\right)\)
\(\Rightarrow2007^{2005}\equiv1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}+2007^{2005}\equiv-1+1=0\left(mod2006\right)\)
\(\Leftrightarrow2005^{2007}+2007^{2005}⋮2006\left(đpcm\right).\)
(21)5×(41)2=6863321781🔥
21^5 x 41^2=6863321781