Tình A=\(\frac{\left(4\cdot7+2\right)\left(6\cdot9+2\right)\left(8\cdot11+2\right)...\left(100\cdot103\right)}{\left(5\cdot8+2\right)\left(7\cdot10+2\right)\left(9\cdot12+2\right)...\left(99\cdot102+2\right)}\)
Những câu hỏi liên quan
Tính\(\frac{\left(4\cdot7+2\right)\left(6\cdot9+2\right)\left(8\cdot11+2\right)..........\left(100\cdot103+2\right)}{\left(5\cdot8+2\right)\left(7\cdot10+2\right)\left(9\cdot12+2\right)..........\left(99\cdot102+2\right)}\)
\(\frac{\left(4\cdot7+2\right)\left(6\cdot6+2\right)\left(8\cdot11+2\right)...\left(100\cdot103+2\right)}{\left(5\cdot8+20\right)\left(7\cdot10+2\right)\left(9\cdot12+2\right)...\left(99\cdot102+2\right)}\)=?
trình bày cách giải hộ em
xem lại đề: không thấy quy luật của tử:
nội suy: Tử là tích=[(2n+2)(2n+5)+2] với n={1...49}
Tuy nhiên mình không thích sửa đề, đề thế nào làm vậy.%
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Tính nhanh :
\(A=\left(1-\frac{2}{6\cdot7}\right)\left(1-\frac{2}{7\cdot8}\right)\left(1-\frac{2}{8\cdot9}\right)\cdot\cdot\cdot\left(1-\frac{2}{51\cdot52}\right)\)
\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot\cdot\cdot\left(1+\frac{1}{99\cdot101}\right)\)
đụ cha mi
mi trù ta thi rớt HK II mà ta giúp mày hả
mấy bài này cũng dễ ẹt nữa
đừng có mơ ta sẽ giúp mày
ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
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\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)
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tìm x
\(\left(x+\frac{1}{2\cdot4}\right)+\left(x+\frac{1}{4\cdot6}\right)+\left(x+\frac{1}{6\cdot8}\right)+\left(x+\frac{1}{8\cdot10}\right)+\left(x+\frac{1}{10\cdot12}\right)=50\frac{5}{24}\)
Tính tổng : a) frac{1}{3cdot5cdot7}+frac{1}{5cdot7cdot9}+frac{1}{7cdot9cdot11}+...+frac{1}{2013cdot2015cdot2017} b) left(1-frac{1}{2^2}right)cdotleft(1-frac{1}{3^2}right)cdotleft(1-frac{1}{4^2}right)cdot...cdotleft(1-frac{1}{2017^2}right) c) left(1-frac{1}{1+2}right)cdotleft(1-frac{1}{1+2+3}right)cdot...cdotleft(1-frac{1}{1+2+3+...+2017}right)
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Tính tổng :
a) \(\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}+\frac{1}{7\cdot9\cdot11}+...+\frac{1}{2013\cdot2015\cdot2017}\)
b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{2017^2}\right)\)
c) \(\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2017}\right)\)
Tính giá trị biểu thức(giút gọn biểu thức)Aleft(left(frac{2}{193}-frac{3}{386}right)cdotfrac{193}{17}+frac{33}{34}right):left(left(frac{7}{2001}+frac{11}{4002}right)cdotfrac{2001}{25}+frac{9}{2}right)Bleft(1+2+3+4+.....+100right)cdotleft(frac{1}{3}-frac{1}{5}+frac{1}{7}-frac{1}{9}right)cdotleft(frac{6}{3}cdot12-2,1cdot3,6right)Cfrac{2cdot8^4cdot27^2+4cdot69}{2^7cdot6^7+2^7cdot40cdot9^4}F1-frac{1}{1+frac{2}{1-frac{3}{1-4}}}ai làm đúng nhanh dễ hiểu thì mk tick cho
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Tính giá trị biểu thức(giút gọn biểu thức)
A=\(\left(\left(\frac{2}{193}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{2001}+\frac{11}{4002}\right)\cdot\frac{2001}{25}+\frac{9}{2}\right)\)
\(B=\left(1+2+3+4+.....+100\right)\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)\cdot\left(\frac{6}{3}\cdot12-2,1\cdot3,6\right)\)
C=\(\frac{2\cdot8^4\cdot27^2+4\cdot69}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(F=1-\frac{1}{1+\frac{2}{1-\frac{3}{1-4}}}\)
ai làm đúng nhanh dễ hiểu thì mk tick cho
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\cdot\left(3n+2\right)}=\frac{n}{6n+4}\)
b.\(\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{\left(4n-1\right)\cdot\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
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b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
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\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)...\left(99\times102+2\right)}=...\)
Tính A:\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)