x.2.( 3.4.5) : 10000
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a)\(\dfrac{1}{10000}+\dfrac{13}{10000}+\dfrac{25}{10000}+...+\dfrac{97}{10000}+\dfrac{109}{10000}\)
\(=\dfrac{1+13+25+...+97+109}{10000}\)
\(=\dfrac{\left(1+109\right)\left[109-1\right]:12+1}{20000}\)
\(=\dfrac{110.10}{20000}=\dfrac{11}{200}\)
b)\(\dfrac{4}{3}\times2019\times0,75\)
=\(\dfrac{4}{3}\times\dfrac{3}{4}\times2019\)
\(=2019\)
c)\(4\times5\times0,25\times\dfrac{1}{5}\times\dfrac{1}{2}\times2\)
\(=\left(4\times\dfrac{1}{4}\right)\left(5\times\dfrac{1}{5}\right)\left(2\times\dfrac{1}{2}\right)\)
\(=1\times1\times1=1\)
Ý d) đặt tính kiểu gì thế ?
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Ở sbt 6 tập mấy ko nhớ có bài tương tự trong ngoặc, mở phần lời giải ra để tính trong ngoặc nha
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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{1998}{1999}\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{1998}{1999}\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{1998}{1999}\)
\(\Leftrightarrow\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1998}{1999}\div\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{3996}{1999}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{1.2}-\frac{3996}{1999}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{-5993}{3998}\)
Như kiểu đề sai hay sao í
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\(M=\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+\frac{37}{1000}+...+\frac{121}{1000}+\frac{133}{1000}\)
\(=\frac{1+13+25+37+...+121+133}{1000}\)
\(=\frac{804}{1000}=\frac{201}{250}\)
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