1+19+3=?
Những câu hỏi liên quan
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)
(1+17)(1+17/2)(1+17/3)...... (1+17/19)
(1+19)(1+19/2)(1+19/3)....(1+19/17)
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)
A= -3/3 - 7/17 + -5/19 + 10/13 + 24/27 - 14/19
B= 1,25 * 7/19 + 5/4 * 15/19 - 1* 1/4 / 19/3
Tính
1/2+1/3+1/4+...1/19+1/20:19/1+18/2+17/3+...+2/18+1/19
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
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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)
Tính Hợp lí nếu có thể!
7/19*-8/11+7/19*-3/11+26/19
1,25*7/19+5/4*15/19 - 1 phần 1/4*3/19
Chứng minh rằng: \(S=\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}< \dfrac{1}{18}\)
`S=1/19+1/19^2+1/19^3+........+1/19^20`
`=>19S=1+1/19+1/19^2+.....+1/19^19`
`=>19S-S=18S=1-1/19^20<1`
`=>S<1/18(đpcm)`
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Giải:
S=\(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\)
19S=\(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\)
19S-S=\(\left(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\right)-\left(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\right)\)
18S=1-\(\dfrac{1}{19^{10}}\)
S=(1-\(\dfrac{1}{19^{10}}\) ):18
S=\(1:18-\dfrac{1}{19^{10}}:18\)
S=\(\dfrac{1}{18}-\dfrac{1}{19^{10}.18}\)
⇒S<\(\dfrac{1}{18}\) (đpcm)
Chúc bạn học tốt!
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S = 119+1192+1193+........+11920S=119+1192+1193+........+11920
⇒ 19S=1+119+1192+.....+11919⇒19S=1+119+1192+.....+11919
⇒ 19S−S=18S=1−11920<1⇒19S-S=18S=1-11920<1
⇒ S<118(đpcm)
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Tinh:
1/19 + 2/18 + 3/17 +...+ 18/2 + 19/1
1/2 + 1/3 + 1/4 +...+ 1/19 + 1/20
\(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\) = \(\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+1\)
= \(\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{20}\)
=\(20.\left(\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}+\frac{1}{20}\right)\)
=\(20.\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}\right)\)
Vì tử số gấp 20 lần mẫu số nên phân số này bằng 20
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\(\frac{\left(1+17\right)+\left(1+\frac{17}{2}\right)+\left(1+\frac{17}{3}\right)+........+\left(1+\frac{17}{19}\right)}{\left(1+19\right)+\left(1+\frac{19}{2}\right)+\left(1+\frac{19}{3}\right)+........+\left(1+\frac{19}{17}\right)}\)
