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Tờ làm luôn, ko ghi đề nữa nhé
\(A=\frac{\frac{24}{12}-\frac{4}{12}+\frac{3}{12}}{\frac{24}{12}+\frac{2}{12}-\frac{3}{12}}\)
\(A=\frac{\frac{23}{12}}{\frac{23}{12}}=1\)
Vậy A=1
\(A=\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)\(=\frac{2-\frac{2}{6}+\frac{2}{8}}{2+\frac{2}{12}-\frac{2}{8}}\)\(=\frac{2\left(1-\frac{1}{6}+\frac{1}{8}\right)}{-2\left(1-\frac{1}{12}+\frac{1}{8}\right)}\)\(=-1\)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
Dat B=1/2+1/2^2+1/2^3+.............+1/2^2012=>2B=1+1/2+1/2^2+.......+1/2^2011
=>2B-B=1/2^2011-1/2^2012=2^2012-2^2011/2^4023
=>A=1+2^2012-2^2011/2^4023
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\)
\(=>B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}=\frac{1\cdot2\cdot3}{2\cdot3\cdot4}=\frac{1}{4}\)
a/ (x-2)/4 =5+x/2
=> \(\frac{x-2}{4}\) = \(\frac{x+5}{2}\)
=> 2(x-2) = 4(x+5)
2x-4 =4x+20
4x-2x=-20-4
2x=-24
x=-12