Tính nhanh:
\(\frac{33.44^2+55^3.33}{45.33^2-99.33^4}\)
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30A=30/2*32+30/3*33+30/4*34=1/2-1/32+1/3-1/33+1/4-1/34=99/100
A=3,3/100
c) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67\cdot67\)
\(=33\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot100=10000\)
a) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67^2\)
\(=\left(33\cdot55+33\cdot45\right)+\left(33\cdot67+67^2\right)\)
\(=33\cdot\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot\left(33+67\right)\)
\(=100\cdot100\)
\(=10000\)
c) Ta có: \(2016\cdot2018-2017^2\)
\(=\left(2017-1\right)\left(2017+1\right)-2017^2\)
\(=2017^2-1-2017^2\)
\(=-1\)
\(A=11\left(\frac{5}{11.6}+\frac{5}{16.21}+......+\frac{5}{36.41}\right)\)
\(=11\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+.....+\frac{1}{36}-\frac{1}{41}\right)\)
\(=11\left(\frac{1}{11}-\frac{1}{41}\right)\)
\(=11.\frac{30}{451}=\frac{30}{41}\)
\(\frac{7}{10}< \frac{6}{7}< \frac{48}{55}< \frac{12}{11}< \frac{8}{7}< \frac{7}{5}< \frac{3}{2}< \frac{9}{4}\)
\(2\frac{17}{20}-1\frac{11}{55}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{6}{5}+\frac{129}{30}x\frac{1}{3}\)
= \(\frac{57}{20}-\frac{24}{20}+\frac{43}{30}\)
= \(\frac{33}{20}+\frac{43}{30}\)
= \(\frac{99}{60}+\frac{86}{60}\)
= \(\frac{37}{12}\)
mk chỉ cần nhìn sơ qua là biết có câu dễ sao bn ko tự nghĩ đi hơi dễ rồi trừ khi bn đố tôi chục câu tiếng anh vật lí văn
\(\frac{33.44^2+55^3.33}{45.33^2-99.33^4}\)
\(=\frac{33.121.16+121.25.33}{5.9.33^2-11.9.33^2.33^2}\)
\(=\frac{33.121\left(16+25\right)}{1089.9.\left(5-11.1089\right)}\)
\(=\frac{3993.41}{9801.\left(-11974\right)}\)
\(=-\frac{163713}{117357174}\).
\(\frac{33.44^2+55^3.33}{45.33^2-99.33^4}\)
=\(\frac{33.\left(4.11\right)^2+\left(5.11\right)^3.33}{9.5.33^2-9.11.33^4}\)
=\(\frac{33.16.11^2+125.11^3.33}{33^2.9.\left(5-11.33^2\right)}\)
= \(\frac{33.11^2.\left(16+125.11\right)}{33^2.9.\left(-11974\right)}\)
= \(\frac{132.1391}{33.9.\left(-107766\right)}\)
= \(\frac{183612}{-32006502}\)
mik ko chắc chắn lắm