\(\frac{1}{2}-\frac{43}{101}+\left(-\frac{1}{3}\right)-\frac{1}{6}\)

\(=\frac{1}{2}-\frac{43}{101}-\frac{1}{3}-\frac{1}{6}\)

\(=\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-\frac{43}{101}\)

\(=0-\frac{43}{101}\)

\(=-\frac{43}{101}\)

P/s tham khảo nhaaaa

-43/101

\(=\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{6}-\frac{43}{101}=\left(\frac{3}{6}-\frac{2}{6}\right)-\frac{1}{6}-\frac{43}{101}\)

\(=\frac{1}{6}-\frac{1}{6}-\frac{43}{101}=0-\frac{43}{101}=\frac{-43}{101}\)

Hok tốt! k mk nha^^

\(M=1-2+3-4+5-6+......+99-100+101\)

\(=\left(1+3+5+.......+99+101\right)-\left(2+4+6+.......+100\right)\)

\(=\left(1+2+3+4+........+100+101\right)-2.\left(2+4+6+......+100\right)\)

\(=\left(1+2+3+4+.......+100+101\right)-\left(1+2+3+.......+50\right)\)

\(=51+52+53+.......+101=\frac{\left(101+51\right).51}{2}=\frac{152.51}{2}=3876\)

m = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10 + ... + 99 - 100 + 101 (có 101 số hạng)
m = (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100) + 101
m= - 1 + (-1) + (-1) + (-1) + ... + (-1) + 101 (có 100 : 2 = 50 số - 1 và 101)
m = - 50 + 101 = (101 - 50)
m

= 51

Ta có M=1-2+3-4+5-6+...+99-100+101

=) M=(1-2)+(3-4)+(5-6)+...+(99-100)+101

=) M=(-1)+(-1)+(-1)+...+(-1)+101

=) M=(-1).50+101        [50 số (-1)]

=) M= -50+101

=) M=51

Giải:

\(\dfrac{1}{2}-\dfrac{43}{100}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)

\(=-\dfrac{43}{100}+\dfrac{1}{2}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)

\(=-\dfrac{43}{100}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)

\(=-\dfrac{43}{100}+\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=-\dfrac{43}{100}+\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

\(=-\dfrac{43}{100}+0=-\dfrac{43}{100}\)

Vậy ...

\(\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}\)

\(=\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)+\left(-\frac{43}{101}\right)\)

\(=\frac{3-2-1}{6}+\frac{-43}{101}\)

\(=0+\frac{-43}{101}\)

\(=-\frac{43}{101}\)

1/2-1/3=1/6 suy ra tổng trên = -43/101

Thanks nhìu

Đặt S = \(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\)

=> 2⁴S = 16S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}\)

=> 16S - S = \(2^3+\frac{1}{2}+\frac{1}{2^5}+...+\frac{1}{2^{97}}-\left(\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+...+\frac{1}{2^{101}}\right)\)

=> 15S = \(2^3-\frac{1}{2^{101}}\)

=> S = \(\frac{2^3-\frac{1}{2^{101}}}{15}\)

Khi đó A = \(\frac{2^3-\frac{1}{2^{101}}}{15}:\left(2^3-\frac{1}{2^{101}}\right)=\frac{1}{15}\)

kết bạn đi toán lớp mấy vậy

toán lớp 7 bạn

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}=\dfrac{100}{101}\)

a: Ta có: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)

\(=\frac{\frac{101.102}{2}}{51}\)

\(=101\)

Đặt S=1³+2³+........+101³

S=(1+2+3+.....+101)²

S=5151²

S=26532801

Đây là công thức mk tự chế nha