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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
\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)
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=\(\frac{4}{13}\)
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
1:
Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)
2) Để A là số nguyên dương thì
\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{6;12\right\}\)
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)