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\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)\times462-2,04\div\left(x+15\right)+11,02=30\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)\times462-2,04\div\left(x+15\right)+11,02=30\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)\times462-2,04\div\left(x+15\right)=30-11,02\)
\(\frac{10}{231}\times462-2,04\div\left(x+15\right)=18,98\)
\(20-2,04\div\left(x+15\right)=18,98\)
\(2,04\div\left(x+15\right)=20-18,98\)
\(2,04\div\left(x+15\right)=1,02\)
\(x+15=2,04\div1,02\)
\(x+15=2\)
\(x=2-15\)
\(x=-13\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{40}{55}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{40}{55}\)
\(x=\frac{55}{55}=1\)
nha.
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
Bài 2: ta có tích riêng thứ nhất là .....5, thứ hai cũng là ....5 -> chữ số tận cùng là: ....5 - ....5 = ...0
Bài 3: Gọi số có hai chữ số đó là ab (a,b =<9)
...........................__..... _
Theo đề bài ta có: ab = 9b
=> b = (2; 3; 4; 5; 6; 7; 8; 9)
..........................................
=> Tương ứng với b ta có ab = (18; 27; 36; 45; 54; 63; 72; 81)
Nhận xét: Chỉ có 45 = 9.5
Vậy số đó là 45
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
\(\frac{1}{11\times16}+\frac{1}{16\times21}+\frac{1}{21\times26}+...+\frac{1}{56\times61}+\frac{1}{61\times66}\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\frac{5}{66}\)
\(=\frac{1}{66}\)
\(\frac{1}{11\times16}+\frac{1}{16\times21}+\frac{1}{21\times26}+...+\frac{1}{56\times61}+\frac{1}{61\times66}\)
\(=\frac{1}{5}\times\left(\frac{5}{11\times16}+\frac{5}{16\times21}+\frac{5}{21\times26}+...+\frac{5}{56\times61}+\frac{5}{61\times66}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\frac{5}{66}=\frac{1}{66}\)