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Đặt A = 1+2-3-4 +........... + 994 - 995 - 996 + 997 + 998
A = 1 + (2-3-4+5) + (6-7-8+9) +............+ (994 - 995 - 996 + 997) + 998
A = 1+ 0 + 0 +.....+0 + 998 = 1 + 998
A = 999
\(\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times\left(\frac{1}{2}-\frac{1}{3}+-\frac{1}{6}\right)\)
\(=\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{10}{99}+\frac{11}{199}-\frac{12}{299}\right)\times0\)
\(=0\)
1 + (-6) + 2 + (-7) + 3 + (-8) + ... + 15 + (-20)
= (1 - 6) + (2 - 7) + (3 - 8)+ ... + (15 - 20)
= -5 + -5 + -5 + ... + -5
= -5 . 15
= -75
( 5x+1 - 61 ) = 24 . 22
( 5x+1 - 61 ) = 26
( 5x+1 - 61 ) = 64
5x+1 = 64 + 61
5x+1 = 125
5x +1 = 53
x + 1 = 3
x = 3 - 1
x = 2
\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
hinh nhu trong sach phat trien lop 6 co thi phai,lau roi quen
\(\frac{x+1}{2017}+\frac{x+2}{2016}=\frac{x+3}{2015}+\frac{x+4}{2014}\)
\(\Leftrightarrow\frac{x+1}{2017}+1+\frac{x+2}{2016}+1=\frac{x+3}{2015}+1+\frac{x+4}{2014}+1\)
\(\Leftrightarrow\frac{x+2018}{2017}+\frac{x+2018}{2016}-\frac{x+2018}{2015}-\frac{x+2018}{2014}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\right)=0\Leftrightarrow x=-2018\)
1+2-3-4+5+6.....+302
=1+(2-3-4+5)+(6-7-8+9)+.......+(299 -300-301-302)
=1 + 0 + 0+ 0 +........+0
= 1
thanks nhung sai rui ket qua la 303 co thay tui noi do