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x,x+1,x+2 chó x là 1 thì bàng 1,1+1,1+2+1,2,3 đúng
b-1,b,b+1 cho b là 1 thì bàng 1-1,1,1+1=0,1,2 đúng
chỉ có 2 câu này đúng thoy
còn 2 câu kia sai oke
với lại chữ E đó là sao vậy
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trường chất lượng cao thì đề toàn bài nâng cao nhé em
còn trường bình thường thì đa số bài khó rơi vào bài cuối cùng
_Muốn vô trw bình thường thì chăm chỉ cần cù là Okê
_Còn vô trw Chất lượng cao bắt buộc phải theo chân lý : Mắt nhìn, tay viết, não phân tích
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Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^2}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)
\(\Rightarrow A=1-\frac{1}{2^7}\)
E= 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256
2E = 2 ( 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256 )
= 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=> E = 2E - E
= (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 )
= 1 - 1/256
= 255/256
k nhá, thanks
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\(\frac{4}{7}:\frac{1}{0,14}+68\%\)
\(=\frac{4}{7}:\frac{100}{14}+\frac{68}{100}=\frac{4}{7}:\frac{50}{7}+\frac{17}{25}\)
\(=\frac{4}{7}\cdot\frac{14}{100}+\frac{68}{100}=\frac{4\cdot14}{7\cdot100}+\frac{17}{25}\)
\(=\frac{1\cdot2}{1\cdot25}+\frac{17}{25}=\frac{2}{25}+\frac{17}{25}\)
\(=\frac{2+17}{25}=\frac{19}{25}\)
\(E=\frac{4}{7}:\frac{1}{0,14}+68\%\)
\(E=\frac{4}{7}:\frac{1}{\frac{7}{50}}+\frac{17}{25}\)
\(E=\frac{4}{7}+\frac{50}{7}+\frac{17}{25}\)
\(E=\frac{57}{7}+\frac{17}{25}=\frac{404}{35}\)
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\(D=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...+\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}......\frac{2018}{2017}.\frac{2019}{2018}\)
\(=\frac{3.4.5....2018.2019}{2.3.4.5....2017.2018}=\frac{2019}{2}\)
\(E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{79.80}+\frac{1}{80.81}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{80}-\frac{1}{81}\)
\(=\frac{1}{4}-\frac{1}{81}\)
\(=\frac{77}{324}\)
\(\text{D}=\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot...\cdot\left(1+\frac{1}{2017}\right)\cdot\left(1+\frac{1}{2018}\right)\)
\(\text{D}=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}\)
\(\text{D}=\frac{2019}{2}\)
\(\text{E}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{79.80}+\frac{1}{80.81}\)
\(\text{E}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{79}-\frac{1}{80}+\frac{1}{80}-\frac{1}{81}\)
\(\text{E}=\frac{1}{4}-\frac{1}{81}=\frac{81}{324}-\frac{4}{324}=\frac{77}{324}\)
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\(E=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}\)
Ta có \(\frac{1}{2.2}>\frac{1}{2.3}\)
\(\frac{1}{3.3}>\frac{1}{3.4}\)
...
\(\frac{1}{49.49}>\frac{1}{49.50}\)
=> \(E=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{49.49}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}=F\)
=> E > F
bằng 2
bằng 2