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2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5
2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5
2\3x-780\11:[13\2.(1\3-1\11)]=-5
2\3x-780\11:[13\2.8\33]=-5
2\3x-780\11:52\33=-5
2\3x-525\13=-5
2\3x=-5+525\13
2\3x=460\13
x=460\13:2\3
x=690\13
1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=15/93
1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3)=15/93
1/2(1/3-1/2x+3)=15/93
=>1/3-1/2x+3=10/31
=>1/2x+3=1/93
=>2x+3=93
2x=93-3=90
=>x=45
Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=90\)
\(x=45\)
Vậy \(x=45\).
A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)
=3/2x4/3x...............x100/99
=2-1/99
=197/99
A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)
A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)
A=\(\frac{100}{2}=50\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%
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=\(\frac{4}{13}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{31}{32}.\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{31}{32}\)
\(\Leftrightarrow\frac{1}{32}=\frac{1}{x+2}\)
\(\Leftrightarrow x+2=32\Rightarrow x=30\)
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)-x=-\frac{100}{99}\)
\(\Rightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)-x=-\frac{100}{99}\)
\(\Rightarrow\left(1-\frac{1}{99}\right)-x=-\frac{100}{99}\)
\(\Rightarrow\frac{98}{99}-x=-\frac{100}{99}\)
\(\Rightarrow x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)
\(\Rightarrow x=\frac{198}{99}=2\)
Vậy x = 2
B = \(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5.1}{1.1.1}=10\)
C = \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}\left(\frac{33}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
câu trả lời này sai rùi nha bn
Cô mik giảng rùi
X= 43
Trần Ngọc Mỹ Anh ara ?! sai đâu ta ?