Bài 1:Rút gọn phân số
a)\(\frac{-2019\cdot2018+1}{\left(-2017\right)\cdot\left(-2019\right)+2018}\)
b)\(\frac{6\cdot9-2\cdot7}{63\cdot3-119}\)
c)\(\frac{2929-101}{2\cdot1919+404}\)
d)\(\frac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}\)
Những câu hỏi liên quan
1.Rút gọn(nếu cần) rồi so sánh
\(\frac{\left(-5\right)^2-5\cdot3^2}{5^3+5^2\cdot3^2}\) ;\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\) và\(\frac{2929-101}{2\cdot1919+404}\)
\(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\cdot\left(x-\frac{1}{2}\right)=\frac{3}{4}\)
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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\(B=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{56^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
Bài 1:Tìm x
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2\cdot x+1\right)\cdot\left(2\cdot x+3\right)}=\frac{9}{19}\)
Bài 2: Tính nhanh
\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2016\cdot2018}\)
ai giúp mình với gấp lắm không có bài là bị phạt đó
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
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\(TínhA=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
Tính A:\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
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)
Đọc tiếp
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) \(\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2017}\right)\)
B =\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\) - \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9+14^3}\)
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(B=\frac{2^{12}.3^6-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(B=\frac{2^{12}.3^4.\left(3^2-1\right)}{2^{12}.3^5.\left(3-1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(B=\frac{8}{6}-\frac{5-35}{9}\)
\(B=\frac{4}{3}+\frac{30}{9}\)
\(B=\frac{4}{3}+\frac{10}{3}\)
\(B=\frac{14}{3}\)
vậy \(B=\frac{14}{3}\)
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