có ai giúp mình với \(\frac{6^3+3\cdot6^2+3^2}{-13}\)
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Giúp mình nhak các friends. Mình mơn nhiều
\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=........................\\\frac{6^3+3\cdot6^2+3^3}{-13}=...................... \)
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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\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
\(\frac{6^3+3.6^2+3}{-13}\)
\(=\frac{216+108+27}{-13}\)
\(=\frac{351}{-13}\)
\(=-27\)
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\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{216+108+27}{-13}\)
\(=\frac{351}{-13}=\frac{27\cdot13}{-13}=-27\)
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\(\frac{6^3+3\cdot6^2\cdot3^3}{-13}\)
\(\frac{6^3+3.6^2.3^3}{-13}\)
\(=\frac{3^3.2^3+3.3^2.2^2.3^3}{-13}\)
\(=\frac{3^3.2^2.\left(2+3^3\right)}{-13}\)
\(=\frac{3^3.2^2.54}{-13}\)
\(=>....\)
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\(\frac{6^3+3\cdot6^2\cdot3^3}{-13}\)
\(=\frac{216+3\cdot36\cdot27}{-13}\)
\(=\frac{216+2916}{-13}\)
\(=\frac{3132}{-13}\)
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Tính nhanh
\(\frac{4\cdot5\cdot6\cdot7}{14\cdot15\cdot16}\)
\(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\)
giúp mình với
\(\frac{4.5.6}{14.15.16}\)=\(\frac{1.1.3}{7.3.4}\)=\(\frac{1.1.1}{7.1.4}\)=\(\frac{1}{28}\)
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\(\frac{4.5.6.7}{14.15.16.17}=\frac{2.2.5.2.3.7}{2.7.3.5.2.2.2.17}=\frac{1}{2.17}=\frac{1}{34}\) ( Gạch bỏ thừa số giống nhau)
\(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}=\frac{2.3.4.5}{3.4.5.6}=\frac{1}{3}\)( Gạch bỏ thừa số giống nhau)
Tíc mình nha!
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\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
đổi tik nha? ^-^
\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
\(=\frac{6^2\cdot\left(6+3\right)+3^3}{-13}\)
\(=\frac{6^2\cdot3^2+3^3}{-13}\)
\(=\frac{3^3\cdot\left(6^2+3\right)}{-13}\)
\(=\frac{3^3\cdot39}{-13}\)
\(=-3^3=-27\)
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Xem thêm câu trả lời
a) \(\frac{\left(-1\right)^3}{15}+\left(-\frac{2}{3}\right):2\frac{2}{3}-\left|-\frac{5}{6}\right|\)
b) \(1\frac{5}{13}-0,\left(3\right)-\left(1\frac{4}{9}+\frac{18}{13}-\frac{1}{3}\right)\)
c) \(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
d) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
Cho S_1-S_2+S_3-S_4+S_5frac{m}{n} với m, n nguyên tố cùng nhau. Biết:S_1frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}S_2frac{1}{2cdot3}+frac{1}{2cdot4}+frac{1}{2cdot5}+frac{1}{2cdot6}+frac{1}{3cdot4}+frac{1}{3cdot5}+frac{1}{3cdot6}+frac{1}{4cdot5}+frac{1}{4cdot6}+frac{1}{5cdot6}S_3frac{1}{2cdot3cdot4}+frac{1}{2cdot3cdot5}+frac{1}{2cdot3cdot6}+frac{1}{2cdot4cdot5}+frac{1}{2cdot4cdot6}+frac{1}{2cdot5cdot6}+frac{1}{3cdot4cdot5}+frac{1}{3cdot4cdot6}+frac{1}{3cdot5cdot6}+frac{1}{4cdot5cdot6}...
Đọc tiếp
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
\(\dfrac{6^2+3\cdot6^2+3^2}{-13}\)
\(=\dfrac{3^2\left(2^2+2^3+1\right)}{-13}=\dfrac{-13\cdot3^2}{13}=-9\)
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