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\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-2-\left(1+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\left(đpcm\right)\)
cau1: y = 7
cau2: số đối của b là 20
( nhìn bài của bn ,mk lại nhớ toi thay tien tai nang, bun wá k mun lam nua)
Câu 1: 7
Câu 2: 20
Câu 3: 1
Câu 4: 100
Câu 5: 20
Câu 6: 7
Câu 7: - 100
Câu 8: 101
Câu 9: 70
Câu 10: Mình quên cách làm mất rồi, bạn thông cảm cho mình nhé!!!
\(A=\dfrac{3^2}{5\cdot14}+\dfrac{3^2}{7\cdot18}+\dfrac{3^2}{9\cdot22}+\dfrac{3^2}{11\cdot26}+\dfrac{3^2}{13\cdot30}\\ =3^2\cdot\left(\dfrac{1}{5\cdot14}+\dfrac{1}{7\cdot18}+\dfrac{1}{9\cdot22}+\dfrac{1}{11\cdot26}+\dfrac{1}{13\cdot30}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}+\dfrac{1}{13\cdot15}\right)\\ =\dfrac{9}{2}\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\right)\\ =\dfrac{9}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)\\ =\dfrac{9}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{15}\right)\\ =\dfrac{9}{4}\cdot\dfrac{2}{15}\\ =\dfrac{3}{10}\)
c: 2|x-3|=500
=>|x-3|=250
=>x-3=250 hoặc x-3=-250
=>x=253 hoặc x=-247
d: x-17=-8+(-17)
=>x-17=-25
hay x=-8
câu 1:20
câu 2:7
câu 3:-10
câu 4:97
câu 5:-20
câu 6: -3
câu 7:686
câu 8:17
câu 9:15 cm
câu 10: 0
b)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\\ 2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\\ 2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\\ B=1-\dfrac{1}{2^{2016}}< 1\)
Vậy B < 1 (đpcm)
a)
Để \(A=\dfrac{3n+2}{n-1}\) nhận giá trị nguyên thì \(3n+2⋮n-1\)
\(3n+2=3n-3+5=3\left(n-1\right)+5\\ 3n+2⋮n-1\Rightarrow3\left(n-1\right)+5⋮n-1\Rightarrow5⋮n-1\Rightarrow n-1\inƯ\left(5\right)\)
Ư(5) = {-5;-1;1;5}