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\(F=\frac{1+\frac{1.2}{2}+\frac{3.4}{2}+...+\frac{100.101}{2}}{1.2+2.3+...+99.100}\)
\(=\frac{1+1.2+3.4+...+100.101}{\left(1.2+2.3+...+99.100\right).2}\)
Tự làm tiếp nhá !
bit lm bài này k giup tui
ta có : ( -5/28 +7/4 + 8/35 ) : (- 69/20)
= ( -25/140 + 245/140 + 32/140 ) x (-20/69)
= (252/140) x (-20/69)
= (9/5) x (-20/69)
= (- 12/23)
tính nhanh:
2 x 3/7 + (2/9 - 10/7) - 5/3 x 9
= 6/7 + 2/9 - 10/7 - 5/3 x 9 = 6/7 + 2/9 - 10/7 - 15
= (6/7 - 10/7 ) + (2/9 - 135/9) = ( - 4/7 ) + (-133/9 )
= (- 36/63) + (-931/63)
= (- 967/63)
Ta gọi A=1.2+2.3+3.4+...+n.(n+1)
3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+n.(n+1)(n+2-n+1)
=[1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)]-[0.1.2+1.2.3+2.3.4+...+(n-1)n(n+1)]
=n(n+1)(n+2)
=> A=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Vậy 1.2+2.3+3.4+...+n(n+1)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
\(\frac{5.\left(2^2.3^2\right)^9\left(2^2\right)^6-2.\left(2^2.3\right)14.3^{ }^4}{22}\)