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1,=0 . [2017/2018+2018/2019]
=>0
2,TH1 x-3=0=>x=3
TH2 y-4=0=>y=4
3, -2/4 = -x/10 = 16/y
=>-1/2 = -x/10 = 16/y
=>-1/2 = -x/10 => -5/10 = -x/10 => x=5
-1/2 = 16/y => 16/-32 = 16/y => y = -32
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
Câu 1:
\(A\in Z\Rightarrow6n-1⋮3n+2\)
\(\Rightarrow6n+4-5⋮3n+2\)
\(\Rightarrow2\left(3n+2\right)-5⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
đến đây tự lm nốt nhé
1. Để A có giá trị nguyên thì \(6n-1⋮3n+2\)
Ta có: \(\left\{{}\begin{matrix}6n-1⋮3n+2\\3n+2⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\2\left(3n+2\right)⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n+4⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n-1+5⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left(6n-1+5\right)-\left(6n-1\right)⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
\(\Rightarrow3n+2\inƯ\left(5\right)\)
\(\Rightarrow3n+2\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow3n\in\left\{-7;\pm3;-1;\right\}\)
\(\Rightarrow n\in\left\{\pm1\right\}\)
Vậy để \(A\in Z\) thì n nhận các giá trị là: \(\pm1\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\\ =\dfrac{5}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\left(1-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\dfrac{100}{101}\\ =\dfrac{250}{101}\)
\(a,\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Vậy \(M=\dfrac{32}{99}\)
b, Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)
\(=1-\dfrac{1}{2012}< 1\) (1)
Do mỗi phân số đều lớn hơn 0 nên \(A>0\) (2)
Từ (1), (2) \(\Rightarrow0< A< 1\)
\(\Rightarrow A\notin N\left(đpcm\right)\)
Vậy...
a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{2}{97}-\dfrac{2}{99}\\ =\dfrac{1}{3}-\dfrac{2}{99}=\dfrac{31}{99}\)
A<\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
A<\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
A<\(\dfrac{1}{2}-\dfrac{1}{2013}\)
A<\(\dfrac{2011}{4026}< 1\)
Vì A bé hơn 1 nên A không phải là số tự nhiên
cau lam ngắn gọn nhỏ
uhm...
sai roi
lan sau lam ngan gon the la ko dung
a) M=\(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + \(\dfrac{2}{7.9}\) + ........+\(\dfrac{2}{97.99}\)
= \(\dfrac{5-3}{3.5}\) + \(\dfrac{7-5}{5.7}\) + \(\dfrac{9-7}{7.9}\) + .........+ \(\dfrac{99-97}{97.99}\)
=\(\dfrac{5}{3.5}\) \(-\) \(\dfrac{3}{3.5}\) + \(\dfrac{7}{5.7}\) \(-\) \(\dfrac{5}{5.7}\) + \(\dfrac{9}{7.9}\) \(-\) \(\dfrac{7}{7.9}\) + .......+ \(\dfrac{99}{97.99}\) \(-\) \(\dfrac{97}{97.99}\)
=\(\dfrac{1}{3}\) \(-\) \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) + .....+ \(\dfrac{1}{97}\) \(-\) \(\dfrac{1}{99}\)
=\(\dfrac{1}{3}\) + \(\dfrac{1}{99}\)
=\(\dfrac{33-1}{99}\)=\(\dfrac{32}{99}\)
b)Cho A =\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) +.......+\(\dfrac{1}{2000^2}\) + \(\dfrac{1}{2011^2}\) + \(\dfrac{1}{2012^2}\)
Ta có: A > 0 (1)
Ta có:\(\dfrac{1}{2^2}\) =\(\dfrac{1}{2.2}\) <\(\dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}\)=\(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}\)=\(\dfrac{1}{4.4}\) < \(\dfrac{1}{3.4}\)
+.......
\(\dfrac{1}{2010^2}\) = \(\dfrac{1}{2010.2010}\) < \(\dfrac{1}{2009.2010}\)
\(\dfrac{1}{2011^2}\) =\(\dfrac{1}{2011.2011}\) <\(\dfrac{1}{2010.2011}\)
\(\dfrac{1}{2012^2}\)=\(\dfrac{1}{2012.2012}\) <\(\dfrac{1}{2011.2012}\)
=>A < \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +......+ \(\dfrac{1}{2009.2010}\) + \(\dfrac{1}{2010.2011}\) + \(\dfrac{1}{2011.2012}\)
=>A < 1\(-\) \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) \(-\) \(\dfrac{1}{4}\) + .....+ \(\dfrac{1}{2009}\)\(-\) \(\dfrac{1}{2010}\) + \(\dfrac{1}{2010}\) \(-\) \(\dfrac{1}{2011}\) + \(\dfrac{1}{2011}\) + \(\dfrac{1}{2012}\)
=> A< 1 \(-\) \(\dfrac{1}{2012}\)
=> A < \(\dfrac{2012-1}{2012}\)
=> A < \(\dfrac{2011}{2012}\) < 1
=> A < 1 (2)
Từ (1) và (2) => 0 < A < 1
Vậy A không phải là một số tự nhiên.
Hay đó! Tự hỏi tự trả lời luôn! Cố gắng phát huy nha!
phan a dau
Phải ấy chứ!ok
sao vậy hỏi còn trả lời làm z
kệ nó vì ko có cái gì làm nên làm lun