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=>1/1 + 1/3 - 1/3 + 1/5 -1/5 + 1/7 - 1/7+..............+ 1/99 - 1/100
= 1/1 - 1/100
=99/100
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{101-99}{99x101}=\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
1-2+3-4+5-6....+97-98+99-100+101
=(1+3+5+7+…+99+101) – (2+4+6+..+89+100)
=(1+101)x51:2 – (100+2)x50:2 = 51
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B=1+2-(3+4)+5+6-..-100+101
B=(3+11+19+...+195)-(7+15+...+199)+101
B=25.99-25.103+101
B=-100+101=1
Vậy B=1
2/1*3+2/3*5+2/5*7...+2/99*101
=(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/99-1/101)
=1-1/101
=100/101
đúng đấy mình làm ùi
Sửa lại
A = 1 + ( 3 - 2 ) + ... + ( 101 - 100 )
A = 1 + 1 + ... + 1
Số số 1 có là :
( 101 - 1 ) : 2 + 1 = 50
A = 1 . 50 + 1
A = 51
\(a,=\dfrac{1}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{1}{3}\times\dfrac{5}{6}=\dfrac{5}{18}\\ b,=\dfrac{4}{5}\times\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{6}=\dfrac{2}{15}\\ c,=456\times99-6\times99+456\\ =456\times\left(99+1\right)-594\\ =456\times100-594=45600-594=45006\\ d,=101\times\left(101-1\right)=101\times100=10100\)
\(\dfrac{3}{1\text{x}3}+\dfrac{3}{3\text{x}5}+...+\dfrac{3}{99\text{x}101}\)
\(=\dfrac{3}{2}\text{x}\left(\dfrac{2}{1\text{x}3}+\dfrac{2}{3\text{x}5}+...+\dfrac{2}{99\text{x}101}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\text{x}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\text{x}\dfrac{100}{101}=\dfrac{150}{101}\)