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Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)
\(=\frac{2.3^2.4^2.5^2...2016^2.2017.2018}{2^2.3^2.4^2.5^2...2017^2}\)
\(=\frac{2018}{2.2017}=\frac{1009}{2017}\)
Nguyễn Tiến Đạt
a)\(|3x-5|=|x+2|\)
=> Ta có 2 trường hợp
*) TH1: 3x-5=x+2
=>3x-x=2+5
=>2x=7
=>x=7:2\(\Rightarrow x=\frac{7}{2}\)
*)TH2: -3x+5=x+2
\(\Rightarrow5-3x=x+2\)
\(\Rightarrow5-2=x+3x\)
\(\Rightarrow3=4x\)
\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)
Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)
A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)
Vậy \(A=\frac{2018}{2019}\)
Chúc bn học tốt
\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)
\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)
\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)
\(=2018.\frac{2}{2019}\)
\(=\frac{4036}{2019}\)
\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)
\(A=\frac{2018}{2019}\)
\(P=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{2017}-1\right)\left(\frac{1}{2018}-1\right)\)
\(P=\left(\frac{-1}{2}\right)\left(\frac{-2}{3}\right)\left(\frac{-3}{4}\right).....\left(\frac{-2016}{2017}\right)\left(\frac{-2017}{2018}\right)\)
\(P=\frac{\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)....\left(-2017\right)}{2.3.4......2017.2018}\)
\(P=\frac{\left(-1\right)\left[\left(-2\right)\left(-3\right)\right]\left[\left(-4\right)\left(-5\right)\right]...\left[\left(-2016\right)\left(-2017\right)\right]}{\left[2.3\right]\left[4.5\right]....\left[2016.2017\right].2018}\)
\(P=\frac{\left(-1\right)\left[2.3\right]\left[4.5\right]....\left[2016.2017\right]}{\left[2.3\right]\left[4.5\right].....\left[2016.2017\right].2018}=\frac{-1}{2018}\)
\(\begin{array}{l}a)A = (2 - \frac{1}{2} - \frac{1}{8}):(1 - \frac{3}{2} - \frac{3}{4})\\ = (\frac{{16}}{8} - \frac{4}{8} - \frac{1}{8}):(\frac{4}{4} - \frac{6}{4} - \frac{3}{4})\\ = \frac{{11}}{8}:\frac{{ - 5}}{4}\\ = \frac{{11}}{8}.\frac{4}{{ - 5}}\\ = \frac{{ - 11}}{{10}}\\b)B = 5 - \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\\ = 5 - \frac{{\frac{3}{3} + \frac{1}{3}}}{{\frac{3}{3} - \frac{1}{3}}}\\ = 5 - \frac{{\frac{4}{3}}}{{\frac{2}{3}}}\\ = 5 - \frac{4}{3}:\frac{2}{3}\\ = 5 - \frac{4}{3}.\frac{3}{2}\\ = 5 - 2\\ = 3\end{array}\)
Chú ý:
Khi thực hiện phép cộng hai phân số, nếu phân số thu được chưa tối giản thì ta rút gọn thành phân số tối giản.
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
\(\frac{2.2017}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2017}}\)
\(=\frac{4043}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2017.2018:2}}\)
\(=\frac{4043}{\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2017.2018}}\)
\(=\frac{4043}{2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)}\)
\(=\frac{4043}{2\left(1+\frac{1}{2}-\frac{1}{2018}\right)}\)
\(=\frac{4043}{2.\frac{1513}{1009}}\)
trog máy tính bấm ra 4034 nưng lại vt 4043 :((