b1:rút gọn
a) \(\frac{10\cdot11+50\cdot55+70\cdot77}{11\cdot12+55\cdot60+77\cdot84}\) b)\(\frac{1\cdot3\cdot5\cdot........\cdot49}{26\cdot27\cdot28\cdot......\cdot50}\)
Những câu hỏi liên quan
1. Chứng minh rằng : Afrac{1}{9}+frac{1}{16}+frac{1}{25}+.......+frac{1}{2500}+frac{1}{4}
2.Rút gọn các phân số sau:
Afrac{10cdot11+50cdot55+70cdot77}{11cdot12+55cdot60+77cdot84}
Bfrac{32768cdot125cdot64cdot81}{8cdot262144cdot81cdot5}
Đọc tiếp
1. Chứng minh rằng : A=\(\frac{1}{9}\)+\(\frac{1}{16}\)+\(\frac{1}{25}\)+.......+\(\frac{1}{2500}\)+\(\frac{1}{4}\)
2.Rút gọn các phân số sau:
A=\(\frac{10\cdot11+50\cdot55+70\cdot77}{11\cdot12+55\cdot60+77\cdot84}\)
B=\(\frac{32768\cdot125\cdot64\cdot81}{8\cdot262144\cdot81\cdot5}\)
Bài 2:
a: \(A=\dfrac{11\cdot10\left(1+5\cdot5+7\cdot7\right)}{11\cdot12\left(1+5\cdot5+7\cdot7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
\(B=\dfrac{1}{8}\cdot\dfrac{125}{5}\cdot\dfrac{81}{81}\cdot\dfrac{64}{8}=25\)
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Cho \(A=1\cdot3\cdot5\cdot7\cdot...\cdot49\)
\(B=\frac{1\cdot2\cdot3\cdot...\cdot48\cdot49\cdot50}{2\cdot4\cdot6\cdot...\cdot48\cdot50}\)
\(C=\frac{26}{2}\cdot\frac{27}{2}\cdot...\cdot\frac{50}{2}\)
So sánh A,B và C
TRẢ LỜI ĐI CÓ DC KO ĐỂ MK CÒN GIẢI
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Giải phương trình
\(\left(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+\cdot\cdot\cdot+\frac{1}{10\cdot60}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\cdot\cdot\cdot+\frac{1}{50\cdot60}\)
Giải phương trình
\(\left(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+\cdot\cdot\cdot+\frac{1}{10\cdot60}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\cdot\cdot\cdot+\frac{1}{50\cdot60}\)
a)A=\(\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}+...+\frac{1}{25\cdot27\cdot29}\)
b)\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1.11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
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Tính \(A=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot5\cdot6}+...+\frac{1}{26\cdot27\cdot28\cdot29\cdot30}\)
A=\(A=\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot25\cdot35}{1\cdot5\cdot7+2\cdot10\cdot19+4\cdot20\cdot28+7\cdot35\cdot49}\)
* là dấu nhân
Cho A=\(1\cdot3\cdot5\cdot7\cdot...\cdot49\)
B=\(\dfrac{26}{2}\cdot\dfrac{27}{2}\cdot...\cdot\dfrac{50}{2}\)
So sanhs A với B
so sánh\(\frac{1\cdot3\cdot5+2\cdot6+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)với \(\frac{303}{708}\)
\(=\frac{219}{520}=\frac{155052}{368160}\)
\(=\frac{303}{708}=\frac{157560}{368160}\)
\(\frac{155052}{368160}< \frac{157560}{368160}\)
VẬY \(\frac{303}{708}\)LỚN HƠN
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