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a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
Bài 3:
a: \(=-8\left(72+19-1\right)=-8\cdot90-720\)
b: \(=-27\left(1011-12-1\right)=-27\cdot998=-26946\)
c: \(=17\cdot\left[29+111\right]+29\cdot\left(-17\right)\)
\(=17\left(29+111-29\right)=17\cdot111=1887\)
d: \(=43\cdot\left(-1\right)+40=-43+40=-3\)
A.\(\dfrac{4}{20}\)+\(\dfrac{16}{42}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)+\(\dfrac{3}{20}\)
=\(\dfrac{1}{5}\)+\(\dfrac{8}{21}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)+\(\dfrac{3}{20}\)
=\((\dfrac{1}{5}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)\()\)+\((\dfrac{8}{21}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)\()\)+\(\dfrac{3}{20}\)
=\((\)\(\dfrac{1}{5}\)+\(\dfrac{2}{5}\)+\(\dfrac{-3}{5}\)\()\)+ 0 +\(\dfrac{3}{20}\)
= 0 + 0 +\(\dfrac{3}{20}\)
=\(\dfrac{3}{20}\)
B.\(\dfrac{42}{46}\)+\(\dfrac{250}{286}\)+\(\dfrac{-2121}{2323}\)+\(\dfrac{-125125}{143143}\)
=\(\dfrac{21}{23}\)+\(\dfrac{125}{143}\)+\(\dfrac{-21}{23}\)+\(\dfrac{-125}{134}\)
=\((\dfrac{21}{23}+\dfrac{-21}{23})+(\dfrac{125}{143}+\dfrac{-125}{134})\)
= 0 + 0
= 0
Chúc bạn Tran Mai học tốt nha