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B=\(\left(1-\dfrac{1}{1+2}\right)\). \(\left(1-\dfrac{1}{1+2+3}\right)\).....\(\left(1-\dfrac{1}{1+2+...+100}\right)\)
B=\(\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{\left(1+100\right)\cdot100:2}\right)\)
B=\(\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{101\cdot100:2-1}{101\cdot100:2}\)
B=\(\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{\left(101.100:2-1\right).2}{101.100}\)
B=\(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}\cdot...\cdot\dfrac{99.102}{100.101}\)
B=\(\dfrac{1.2.3.4.....99}{3.4.5....100}.\dfrac{4.5.6.....102}{3.4.5.....101}\)
B=\(\dfrac{2}{100}\).\(\dfrac{102}{3}\)
B=\(\dfrac{17}{25}\)
a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
Ta có S = ( 1/2 - 1) : ( 1/3 - 1) : (1/4 - 1) :... : ( 1/50 - 1)
S = -1/2 : ( -2/3) : ( -3/4) : ... : ( -49/ 50)
S= -1/2 x (-3/2) x ( -4/3) x ... x (-50/49)
S= -1/2 x 1/3 x 50
S= -25/3
\(Q=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(Q=\frac{1}{5}-\frac{1}{200}=\frac{39}{200}\)
\(Q=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
a) \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right].\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right).\left(-3\right)-25\)
\(=\left(-7\right)-120-25\)
\(=-152\)
b) \(\left(-2\right)^3.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=\left(-8\right).3-\left(1+8\right):9\)
\(=\left(-24\right)-9:9\)
\(=\left(-24\right)-1\)
\(=-25\)
Bài giải
a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)
\(=-7-120-25\)
\(=-127-25\)
\(=-152\)
b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)
\(=-8\cdot3-\left(1+8\right)\text{ : }9\)
\(=-24-9\text{ : }9\)
\(=-24-1\)
\(=-25\)
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-2-\left(1+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\left(đpcm\right)\)
\(P=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=\frac{2^2-2}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{99.101}{100^2}\)
\(=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(=\frac{1}{100}.\frac{101}{2}\)
\(=\frac{101}{200}\)
\(P=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)......\left(1-\frac{1}{100^2}\right)\)
\(\Rightarrow P=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}......\frac{100^2-1}{100^2}\)
\(\Rightarrow P=\frac{3}{2^2}.\frac{8}{3^2}...............\frac{9999}{100^2}\)
\(\Rightarrow P=\frac{1.3}{2^2}.\frac{2.4}{3^2}............\frac{99.101}{100^2}\)
\(\Rightarrow P=\frac{\left(1.2............99\right).\left(3.4............101\right)}{\left(2.3..............100\right).\left(2.3...........100\right)}\)
\(\Rightarrow P=\frac{1.101}{100.2}=\frac{101}{200}\)
Vậy \(P=\frac{101}{200}\)
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