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cho công thức tổng quát nè (do tui tự nghĩ ra đó :))
\(\left(a+b\right)\left(\frac{1}{?}\right)=\frac{a+b}{?}\) dựa vô đây nhân (1+17) và (1+19) và từng cái ngoặc kia là đc
\(A=\frac{\frac{1+17}{1}\cdot\frac{2+17}{2}\cdot\frac{3+17}{3}\cdot...\cdot\frac{19+17}{19}}{\frac{1+19}{1}\cdot\frac{2+19}{2}\cdot\frac{3+19}{3}\cdot...\cdot\frac{17+19}{17}}=\frac{18\cdot19\cdot20\cdot...\cdot36}{1\cdot2\cdot3\cdot...\cdot19}:\frac{20\cdot21\cdot22\cdot...\cdot36}{1\cdot2\cdot3\cdot...\cdot17}\)
\(=\frac{18\cdot19}{18\cdot19}=1\)
a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
\(Giải:\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.......+\frac{1}{20}\left(1+.......+20\right)\)
\(B=1+\frac{1}{2}\left(\frac{3.2}{2}\right)+\frac{1}{3}\left(\frac{4.3}{2}\right)+\frac{1}{4}\left(\frac{5.4}{2}\right)+.......+\frac{1}{20}\left(\frac{21.20}{2}\right)\)
\(B=1+\frac{3.2}{2.2}+\frac{4.3}{3.2}+\frac{5.4}{4.2}+.........+\frac{21.20}{20.2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+.........+\frac{21}{2}=\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+.........+\frac{21}{2}-\frac{1}{2}\)
\(=\frac{1+2+3+.........+21-1}{2}=\frac{22.21-1}{4}=\frac{242-1}{4}=\frac{241}{4}=60\frac{1}{4}\)
Cố gắng lên, mấy thánh sẽ phù hộ con!!!
thánh rảnh thì thánh giải hộ con với