tính giá trị biểu thức
\(\left(-1\frac{1}{5}\right).\left(-1\frac{1}{6}\right).\left(-1\frac{1}{7}\right)\).\(\left(-1\frac{1}{8}\right).\left(-1\frac{1}{9}\right).\left(-1\frac{1}{10}\right)\)
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Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)
\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)
A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)
Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{10}{x^2+12x+11}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)
\(P=\frac{37}{60}\)
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)
\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)
\(Q=\frac{1044}{4375}\)
=\(-\frac{6}{5}\).\(\frac{-7}{6}\).\(\frac{-8}{7}\).\(\frac{-9}{8}\).\(\frac{-10}{9}\).\(\frac{-11}{10}\)
=\(\frac{7}{5}\).\(\frac{9}{7}\).\(\frac{11}{9}\)
=\(\frac{11}{5}\)
\(=\frac{-6}{5}\times\frac{-7}{6}\times\frac{-8}{7}\times\frac{-9}{8}\times\frac{-10}{9}\times\frac{-11}{10}\)
\(=\frac{\left(-6\right).\left(-7\right).\left(-8\right).\left(-9\right).\left(-10\right).\left(-11\right)}{5.6.7.8.9.10}\)
\(=\frac{6\times7\times8\times9\times10\times11}{5\times6\times7\times8\times9\times10}\)
Triệt tiêu các thừa số bằng nhau ở tử và mẫu, ta có kết quả là \(\frac{11}{5}\)