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\(=3\cdot2^3\cdot5^2+2\cdot2^6\cdot2^2\cdot3+2^3\cdot2\cdot3\cdot2^2+3\cdot2\cdot3\cdot2^2\\ =2^3\cdot3\cdot5^2+2^9\cdot3+2^6\cdot3+2^3\cdot3^2\\ =2^3\cdot3\left(5^2+2^6+2^3+3\right)=24\left(25+64+8+3\right)\\ =24\cdot100=2400\)
a, \(x:\left[\left(1800+600\right):30\right]=560:\left(315-35\right)\)
\(\Rightarrow\) \(x:\left[2400:30\right]=560:280\)
\(\Rightarrow\) \(x:80=2\)
\(\Rightarrow\) \(x=160\)
b, \(\left[\left(250-25\right):15\right]:x=\left(450-60\right):130\)
\(\Rightarrow\) \(\left[225:15\right]:x=390:130\)
\(\Rightarrow\) \(15:x=3\)
\(\Rightarrow\) \(x=5\)
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a) 24.5 - [ 131. ( 13 - 4 )2 ]
=120 - [ 131 . 92 ]
=120 - [ 131 . 81 ]
=120 - 10611
= - 10491
b) 100 : {230:[450−(4−53−52.25)]}
= 100 : \(\left\{230:\left[450-\left(4-125-25.25\right)\right]\right\}\)
= \(100:\left\{230:\left[450-\left(4-125-625\right)\right]\right\}\)
= \(100:\left\{230:\left[450-\left(-746\right)\right]\right\}\)
=\(100:\left\{230:1196\right\}\)
= 100 : \(\dfrac{5}{26}\)= 520
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)
\(Q=\frac{1}{100}\)
\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)
\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)
\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)
Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới
\(P=\frac{201}{100}\)