(1+17)(1+17/2)(1+17/3)...... (1+17/19)
(1+19)(1+19/2)(1+19/3)....(1+19/17)
Những câu hỏi liên quan
Tính giá trị biểu thức: A =
(1+17).(1+17/2).(1+17/3).(1+17/4)...(1+17/19) phần (1+19).(1+19/2).(1+19/3).(1+19/4)...(1+19/17)
Tính:
1)1+1/2.(1+2)+1/3.(1+2+3)+...+1/100.(1+2+3+...+100)
2)(1+17)(1+17/2)(1+17/3)...(1+17/19) / (1+19)(1+19/2)(1+19/3)...(1+19/17)
Tinhs giá trị của biểu thức sau
(1+17)x(1+17/2)x(1+ 17 / 3) x .......x(1+17/19)
(1+19)x(1+19/2)x(1+19/3)x.....x(1+19/17)
\(\frac{\left(1+17\right)+\left(1+\frac{17}{2}\right)+\left(1+\frac{17}{3}\right)+........+\left(1+\frac{17}{19}\right)}{\left(1+19\right)+\left(1+\frac{19}{2}\right)+\left(1+\frac{19}{3}\right)+........+\left(1+\frac{19}{17}\right)}\)
Tinh: B= (1+17)x(1+17/2)x(1+17/3)....(1+17/19)
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(1+19)x(1+19/2)x(1+19/3)....(1+19/17)
\(A=\frac{\left(1+17\right)\left(1+\frac{17}{2}\right)\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right)\left(1+\frac{19}{2}\right)\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)=?
\(A=\frac{18.\frac{19}{2}.\frac{20}{3}....\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{\frac{18.19.20...36}{1.2.3...17.18.19}}{\frac{20.21.22....36}{1.2.3...17}}=\frac{18.19.\left(20.21...36\right)}{\left(1.2.3...17\right).18.19}.\frac{1.2.3...17}{20.21.22....36}=1\)
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Tính
\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)....\left(1+\frac{19}{17}\right)}\)
\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)...\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right)}\)
\(=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{18.19.20...36}{1.2.3...19}:\frac{20.21.22...36}{1.2.3...17}\)
\(=\frac{18.19.20...36}{1.2.3...19}.\frac{1.2.3...17}{20.21.22....36}=\frac{1.2.3...17.18...36}{1.2.3...19.20...36}=1\)
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Tính giá trị biểu thức: A=\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right).}\)
Tính \(A=\frac{\left(1+17\right)+\left(1+\frac{17}{2}\right)+\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+19\right)\left(1+\frac{19}{2}\right)\left(1+\frac{19}{3}\right)....\left(1+\frac{19}{17}\right)}\)