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a: =>3/7x=5/7-1=-2/7
hay x=-2/3
b: =>x+3/4=14/25
=>x=14/25-3/4=-19/100
c: =>-1/4-x=-7/5
=>x+1/4=7/5
hay x=7/5-1/4=23/20
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g: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{4}{9}-\dfrac{1}{3}=\dfrac{1}{9}\)
=>x-1/2=1/3 hoặc x-1/2=-1/3
=>x=5/6 hoặc x=-1/6
h: \(\Leftrightarrow x\cdot\dfrac{16}{5}-\dfrac{22}{15}:\dfrac{11}{3}=\dfrac{7}{20}\)
\(\Leftrightarrow x\cdot\dfrac{16}{5}=\dfrac{7}{20}+\dfrac{22}{15}\cdot\dfrac{3}{11}=\dfrac{7}{20}+\dfrac{2}{5}=\dfrac{3}{4}\)
hay x=15/64
i: \(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)=\dfrac{13}{2}:\dfrac{65}{61}=\dfrac{61}{10}\)
=>2x=-8/5
hay x=-4/5
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a: \(=\dfrac{2}{7}\cdot\dfrac{-3}{4}\cdot\dfrac{4}{7}=\dfrac{-6}{49}\)
b: \(=\dfrac{3}{5}:\dfrac{1}{15}=\dfrac{3}{5}\cdot15=9\)
c: \(=\dfrac{41}{7}-3-\dfrac{1}{2}=\dfrac{33}{14}\)
d: \(=-\dfrac{25}{12}-\dfrac{23}{12}-\dfrac{3}{2}=-4-\dfrac{3}{2}=-\dfrac{11}{2}\)
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TRẢ LỜI:
Đáp án: C
Trong hiện tượng giao thoa sóng trên mặt nước, khoảng cách giữa hai cực đại liên tiếp nằm trên đường nối tâm hai sóng có độ dài là một nửa bước sóng.
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\(I=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2009.2010}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}\\ =\dfrac{2009}{2010}\)
\(K=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{2008.2010}\\ =2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\\ =2.\dfrac{502}{1005}\\ =\dfrac{1004}{1005}\)
\(F=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}.\dfrac{10}{33}\\ =\dfrac{10}{99}\)
\(I=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}=\dfrac{2010-1}{2010}=\dfrac{2008}{2010}\)
\(K=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}=\dfrac{502}{1005}\)
\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{99.100}=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{100}\right)=\dfrac{97}{900}\)
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Những số có chữ số tận cùng là 2,4,8 khi nâng lên mũ 4 có tận cùng là 6
Thật vậy
\(4^{2k}=2^{4k}=...6\)
\(4^{2k+1}=2^{4k+2}=2^{4k}.4=\left(...6\right).4=...4\)
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2021+(-247)+(-53)-2021=(2021-2021)+[(-247)+(-53)]
=0+(-300)
=-300
d) \(x+\dfrac{3}{4}=\dfrac{36}{144}.\dfrac{-12}{9}\)
\(\Rightarrow x+\dfrac{3}{4}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{-4}{12}-\dfrac{9}{12}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
e)
\(\dfrac{8}{23}.\dfrac{46}{24}=\dfrac{1}{3}.x\)
\(\Rightarrow\dfrac{1}{3}.x=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}.3\)
\(\Rightarrow x=2\)
f)
\(\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{1}{7}\)
\(\dfrac{1}{5}:x=\dfrac{7}{35}-\dfrac{5}{35}\)
\(\dfrac{1}{5}:x=\dfrac{2}{35}\)
\(\Rightarrow x=\dfrac{1}{5}:\dfrac{2}{35}\)
\(\Rightarrow x=\dfrac{1}{5}.\dfrac{35}{2}\)
\(\Rightarrow x=\dfrac{7}{2}\)
f) 1/5 : x = 1/5 - 1/7
1/5 : x = 2/35
x = 1/5 : 2/35
x = 7/2
e) x + 3/4 = 36/144 . -12/9
x + 3/4 = -1/3
x = -1/3 - 3/4
x = -13/12