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\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
a) \(\left(2\frac{5}{6}+1\frac{4}{9}\right):\left(10\frac{1}{12}-9\frac{1}{2}\right)\)
\(=\left(\frac{17}{6}+\frac{13}{9}\right):\left(\frac{121}{12}-\frac{19}{2}\right)\)
\(=\frac{77}{18}:\frac{7}{12}\)
\(=\frac{22}{3}\)
b) \(1\frac{5}{18}-\frac{5}{18}.\left(\frac{1}{15}+1\frac{1}{12}\right)\)
\(=\frac{23}{18}-\frac{5}{18}.\left(\frac{1}{15}+\frac{13}{12}\right)\)
\(=\frac{23}{18}-\frac{5}{18}.\frac{23}{20}\)
\(=\frac{23}{18}-\frac{23}{72}\)
\(=\frac{23}{24}\)
c) \(-1\frac{1}{7}.\left(9\frac{1}{2}-8,75\right):\frac{2}{7}+0,625:1\frac{2}{3}\)
\(=\frac{-8}{7}.\left(\frac{19}{2}-\frac{35}{4}\right):\frac{2}{7}+\frac{5}{8}:\frac{5}{3}\)
\(=\frac{-8}{7}.\frac{3}{4}:\frac{2}{7}+\frac{3}{8}\)
\(=\frac{-8}{7}.\frac{3}{4}.\frac{7}{2}+\frac{3}{8}\)
\(=-3+\frac{3}{8}\)
\(=\frac{-21}{8}\)
Chúc bn học tốt !!
Ta có: \(D=2016\left(1-\frac{2}{3}\right)\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2017}\right)\)
\(=2016.\frac{1}{3}.\frac{3}{5}.\frac{5}{7}...\frac{2015}{2017}\)\(=2016.\left(\frac{1}{3}.\frac{3}{5}.\frac{5}{7}...\frac{2015}{2017}\right)\)
\(=2016\left(\frac{1.3.5.7...2015}{3.5.7....2015.2017}\right)\)\(=2016.\frac{1}{2017}=\frac{2016}{2017}\)
Vậy \(D=\frac{2016}{2017}\)
d, \(\frac{1023}{2^1+2^2+...+2^{10}}\)
\(\text{Đặt}:S=2^1+2^2+...+2^{10}\)
\(2S=2.\left(2^1+2^2+..+2^{10}\right)\)
\(2S=2^2+2^3+..+2^{11}\)
\(S=2S-S=\left(2^2+2^3+...+2^{11}\right)-\left(2^1+2^2+...+2^{10}\right)\)
\(S=2^{11}-2^1=2^{11}-1\)
Thay S vào biểu thức \(\frac{1023}{2^1+2^2+...+2^{10}}\),ta được
\(\frac{1023}{2^{11}-1}=\frac{1023}{2047}\)
Vậy ......