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\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{4}{15}\)
\(=\frac{2}{15}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{1}{13.15}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\)4/15
=2/15
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{1}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\div2\)
\(\Rightarrow A=\frac{50}{101}\)
\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)
\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)
\(2C=\dfrac{4}{13}\)
\(C=\dfrac{2}{13}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\frac{12}{39}\)
\(C=\frac{4}{26}=\frac{2}{13}\)
Q = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
Q = \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{2013.2015}\right)\)
Q = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2015}\right)\)
Q = \(\frac{1}{2}.\frac{2012}{6045}=\frac{1002}{6045}\)
\(Q=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\)
\(\Rightarrow Q.2=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\right)\)
\(\Rightarrow Q.2=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2013.2015}\)
\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{2015}\)
\(\Rightarrow Q.2=\frac{2012}{6045}\)
\(\Rightarrow Q=\frac{2012}{6045}.\frac{1}{2}=\frac{1006}{6045}\)
Mk tinh nhẩm, nên ko bt kết quả có đúng ko
nên bn thử tính lại kết quả nha!!!
Chúc bn hok tốt!!!
\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(M=\frac{1}{3}-\frac{1}{13}\)
\(M=\frac{10}{39}\)
\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(M=\frac{1}{2}.\frac{10}{39}\)
\(M=\frac{5}{39}\)
tk mk nha bn
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{87.89}\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{87}-\frac{1}{89}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{89}\right)\)
= \(\frac{1}{2}.\frac{86}{267}=\frac{43}{267}\)
~~~
Đáp số to quá, tớ không chắc là mình đúng đâu.
#Sunrise
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\cdot\frac{4}{15}=\frac{2}{15}\)
Gọi biểu thức trên là A.
Ta có : 2A = \(\frac{2}{3.5}+\frac{2}{5.7}+................+\frac{2}{89.91}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.............+\frac{1}{89}-\frac{1}{91}\)
=> 2A = \(\frac{1}{3}-\frac{1}{91}=\frac{88}{273}\)
=> A = \(\frac{88}{273}\div2=\frac{44}{273}\)
k cho mình nha bạn
Goi tong tren la : A
Ta co: A = 1/3.5 + 1/5.7 + 1/7.9 + ......+ 1/89.91
2A = 2/3.5 + 2/5.7 +......+ 2/89.91
2A = 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/89 - 1/91
2A = 1/3 - 1/91
2A = 88/273
A = 44/273