\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
Những câu hỏi liên quan
\(\frac{6^3+3\cdot6^2\cdot3^3}{-13}\)
\(\frac{6^3+3.6^2.3^3}{-13}\)
\(=\frac{3^3.2^3+3.3^2.2^2.3^3}{-13}\)
\(=\frac{3^3.2^2.\left(2+3^3\right)}{-13}\)
\(=\frac{3^3.2^2.54}{-13}\)
\(=>....\)
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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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\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
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
có ai giúp mình với \(\frac{6^3+3\cdot6^2+3^2}{-13}\)
\(\frac{6^3+3.6^2+3^2}{-13}\)
\(=\frac{\left(2.3\right)^3+3.\left(2.3\right)^2+3^2}{-13}\)
\(=\frac{2^3.3^3+\left(3.3^2\right).2+3^2}{-13}\)
\(=\frac{3^2.\left(2^3.3+3.2+1\right)}{-13}\)
\(=\frac{3^2.31}{-13}\)
\(=\frac{9.31}{-13}\)
\(=\frac{273}{-13}\)
\(=-21\)
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Giúp mình nhak các friends. Mình mơn nhiều
\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=........................\\\frac{6^3+3\cdot6^2+3^3}{-13}=...................... \)
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) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
Tính:
a,\(\frac{4^2\cdot4^3}{2^{10}}\)
b,\(\frac{0,6^5}{0,2^6}\)
c,\(\frac{2^7\cdot9^3}{6^5\cdot8^2}\)
d,\(\frac{6^3+3\cdot6^2+3^3}{-13}\)
a) \(\frac{4^2.4^3}{2^{10}}\)\
\(=\frac{4^5}{\left(2^2\right)^{10}}=\frac{4^5}{4^{10}}=\frac{1}{4^5}\)
b) \(\frac{0,6^5}{0,2^6}=\frac{\left(0,2.3\right)^5}{0,2^6}=\frac{0,2^5.3^5}{0,2^6}=\frac{3^5}{0,2}=5.3^5\)
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\(\dfrac{6^2+3\cdot6^2+3^2}{-13}\)
\(=\dfrac{3^2\left(2^2+2^3+1\right)}{-13}=\dfrac{-13\cdot3^2}{13}=-9\)
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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 \(m+n\)
\(\dfrac{6^3+3\cdot6^2+3^3}{-13}\)
\(\dfrac{6^3+3.6^2+3^3}{-13}\)=\(\dfrac{3^3.2^3+3.2^2.3^2+3^3}{-13}\)=\(\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)
=\(\dfrac{3^3.13}{-13}\)=\(-3^3\)=\(-27\)
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