-2/3+4/3.5+4/5.7+4/7.9+....+4/97.99+101/99
B=-2/3+4/3.5+4/5.7+4/7.9+.......+4/97.99+101/99
4/3.5 + 4/5.7 + 4/7.9 + ... + 4 /97.99 = ?
\(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+...+\frac{4}{97\cdot99}\)
\(=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+...+\frac{2\cdot2}{97\cdot99}\)
\(=\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+\frac{2}{7}-\frac{2}{7}+\frac{2}{9}-...+\frac{2}{97}-\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{99}\)
\(=\frac{64}{99}\)
Tính giá trị biêut hức;B=2/1.3-4/3.5+6/5.7-8/7.9+...-96/95.97+98/97.99
các bạn cho mk hỏi câu này
2/3.5+2/5.7+2/7.9+...+2/97.99
thì mk sẽ viết thành
1/3.5+1/5.7+1/7.9+...+1/97.99
hay
2.(1/3.5+1/5.7+1/7.9+...+1/97.99)
giúp mk với
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99
Tính:
a) M=2/3.5+2/5.7+2/7.9+...+2/97.99
b) N=3/5.7+3/7.9+3/9.11+...+3/197.199
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
Tìm x, biết: 1/3 + 1/3.5 + 1/5.7 + 1/7.9 +...+ 1/(x+2).(x+4)=50/101
<=> 2/1.3 + 2/3.5 + 2/5.7 +....+ 2/(x+2)(x+4) = 100/101
<=> 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.....+ 1/x+2 - 1/x+4 = 100/101
<=> 1 - 1/x+4 = 100/101
<=> 1/x+4 = 1 - 100/101 <=> 1/x+4 = 1/101 <=> x+4 = 101 <=> x= 101 - 4 = 97
:)
Cho A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 +.....+ 2/97.99 và B = 1^2 / 1.2 x 2^2/2.3 x 3^2 / 3.4 x 4^2 /4.5 x .... x 98^2 / 98.99. Chứng tỏ A = 98B
Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)
Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)
=> A = 98B
các bạn có về sweet home
Chứng tỏ:
\(\frac{4}{3.5}\)+ \(\frac{4}{5.7}\)+ \(\frac{4}{7.9}\)+.................+ \(\frac{4}{97.99}\)> 65%
ta co : 65%=0,65
goi A= 4.(1/3.5+1/5.7+1/7.9+............+1/97.99)
2A=4.( 2/3.5+2/5.7+2/7.9+...............+2/97.99)
2A=4.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)
2A=4.(1/3-1/99)
2A=4.(33/=99+1/99)
2A=4.34/99
2A=136/99
A=136/99:2
A=68/99=0,69=0,68
Vi A=0,68 > 0,65
=> A > 65%
Tìm giá trị của biểu thức \(P=\frac{2}{1.3}-\frac{4}{3.5}+\frac{6}{5.7}+\frac{8}{7.9}+...-\frac{96}{95.97}+\frac{98}{97.99}\)