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\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-1\frac{3}{4}\)
\(=2\frac{1}{4}\)
\(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=4-\frac{7}{4}\)
\(=\frac{16}{4}-\frac{7}{4}\)
\(=\frac{9}{4}\)
= \(6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}=6\frac{5}{7}-2\frac{5}{7}\)\(-1\frac{3}{4}\)= \(4-1\frac{3}{4}\)=\(\frac{9}{4}\)
47/7 - (7/4 + 19/7) =47/7 -125/28 =gomen =tao xin lỗi chịu rùi = <3
\(M=1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)
đặt \(A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)
\(3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{18}}-\frac{1}{3^{19}}\)
\(4A=1-\frac{1}{3^{20}}\)
\(A=\frac{1-\frac{1}{3^{20}}}{4}\)
\(M=1+\frac{1-\frac{1}{3^{20}}}{4}=\frac{5-\frac{1}{3^{20}}}{4}\)
Ta có : 1:M=1+3-3^2+3^3-3^4+....+3^19-3^20
1/M=(1+3^2+3^4+....3^20)-(3+3^3+..+3^19)
1/M=[(3^20-1)/8]-[(3^21-3)/8]
1/M=[3^20-3^21+(-2)]/8
Bạn tự làm tiếp nhé
\(\frac{\frac{3}{11}-\frac{3}{13}+\frac{3}{17}-\frac{3}{19}}{\frac{4}{11}-\frac{4}{13}+\frac{4}{17}-\frac{4}{19}}=\frac{3\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}{4\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}=\frac{3}{4}\)
\(\frac{3.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}{4.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}\) = \(\frac{3}{4}\)
--------------------- Hok Tốt-----------------------------
Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)
Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)
1/
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)
\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=50-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)
2/
\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)