Tính S=(4/1.3)-(8/3.5)+(12/5.7)-(16/7.9)+(20/9.11)-(24/11.13)+(28/13.15)
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\(-\frac{2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-\frac{2}{9.11}-\frac{2}{11.13}-\frac{2}{13.15}\)
\(=\left(-\frac{2}{1.3}\right)+\left(-\frac{2}{3.5}\right)+\left(-\frac{2}{5.7}\right)+\left(-\frac{2}{7.9}\right)+\left(-\frac{2}{9.11}\right)+\left(-\frac{2}{11.13}\right)+\left(-\frac{2}{13.15}\right)\)
\(=\left(-2\right).\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{15}\right)=\left(-2\right).\frac{14}{15}\)
\(=-\frac{28}{15}\)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
\(\frac{x}{3.5}+\frac{x}{5.7}+\frac{x}{7.9}+...+\frac{x}{13.15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\frac{4}{15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}=\frac{4}{45}:\frac{4}{15}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}.2\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
_Chúc bạn học tốt_
1) P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15
P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)
P=1/3-1/15= 4/15
2) a/ 0,2:1+3/5+80%
= 2/10:8/5+8/10
= 2/10.5/8+8/10
= 1/8 + 4/5 = 5/40 + 32/40 = 37/40
b/ 0,5:5/4-2+1/5
= 5/10:5/4-11/5
= 5/10.4/5-11/5
=2/5-11/5 = -9/5
1)
2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10
=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)
= (2/3-2/15)+ [2(1-1/10)]
=8/15+9/5
=7/3
2)
11/12+11/12.24+...+11/88.99
=11-1/9
=10/8/9
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{13}-\dfrac{1}{15}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)
Chúc bạn học tốt!!!
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
= \(\dfrac{1}{3}-\dfrac{1}{15}\)
= \(\dfrac{4}{15}\)
Ta có: \(N=\dfrac{5}{3.5}-\dfrac{5}{5.7}-\dfrac{5}{7.9}-\dfrac{5}{9.11}-\dfrac{5}{11.13}-\dfrac{5}{13.15}
\)
\(\Rightarrow N=5\left(\dfrac{1}{3.5}-\dfrac{1}{5.7}-\dfrac{1}{7.9}-\dfrac{1}{9.11}-\dfrac{1}{11.13}-\dfrac{1}{13.15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{13}-\dfrac{1}{13}+\dfrac{1}{15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{15}\right)=\dfrac{5}{2}\left(\dfrac{5+1}{15}\right)=\dfrac{5}{2}.\dfrac{6}{15}\)
\(\Rightarrow N=\dfrac{3}{3}=1\)
\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
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