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a, Ta có :\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\\ \Rightarrow2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\\ \Rightarrow2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\right)\\ \Rightarrow A=1-\dfrac{1}{2^{50}}< 1\\ \Rightarrow A< 1\) Vậy \(A< 1\)
b, Ta có :
\(B=\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}\)Vậy \(B< \dfrac{1}{2}\)
c, Ta có :
\(C=\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\\ \Rightarrow4C=1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\\\Rightarrow4C-C=\left(1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\right)-\left(\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\right)\\ \Rightarrow3C=1-\dfrac{1}{4^{1000}}< 1\\ \Rightarrow C< \dfrac{1}{3}\)Vậy \(C< \dfrac{1}{3}\)
a) \(P=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+...+\frac{9+10}{9^2.10^2}\)
\(P=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\) ( rút gọn số mũ nhé )
\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)
\(P=1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
Vì \(\frac{9}{10}< 1\Rightarrow P< 1\) (đpcm)
b) Chút nữa mình làm nhé ^^
b)
\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
Ta so sánh giữa A và Q.
\(\frac{1}{1.2}>\frac{1}{3};\frac{1}{2.3}>\frac{1}{3^2};\frac{1}{3.4}>\frac{1}{3^3};....;\frac{1}{100.101}>\frac{1}{3^{100}}\)
\(\Rightarrow Q< A\)
Ta lại tiếp tục so sánh A và \(\frac{1}{2}\)
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\Leftrightarrow A< \frac{1}{2}\)
Ta được:
\(Q< A< \frac{1}{2}\Leftrightarrow Q< \frac{1}{2}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)....................\left(\dfrac{1}{10^2}-1\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right)...........\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-99}{100}\)
\(=\dfrac{\left(-1\right).3}{2.2}.\dfrac{\left(-2\right).4}{3.3}..................\dfrac{\left(-9\right).11}{10.10}\)
\(=\dfrac{\left(-1\right)\left(-2\right)..........\left(-9\right)}{2.3.....10}.\dfrac{3.4....11}{2.3....10}\)
\(=\dfrac{-1}{10}.\dfrac{11}{2}\)
\(=\dfrac{-11}{20}< \dfrac{-10}{20}=\dfrac{-1}{2}\)
\(\Leftrightarrow A< \dfrac{-1}{2}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{10^2}-1\right)\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right).....\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right)....\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.....\dfrac{-99}{100}\)
\(A=\dfrac{\left(-1\right).3}{4}.\dfrac{\left(-1\right).8}{9}......\dfrac{\left(-1\right).99}{100}\)
\(A=\dfrac{\left(-1\right).1.3}{2.2}.\dfrac{-1.2.4}{3.3}....\dfrac{-1.9.11}{10.10}\)
\(A=\dfrac{-1.3}{2.2}.\dfrac{-2.4}{3.3}....\dfrac{-9.11}{10.10}\)
\(A=\dfrac{\left(-1\right)\left(-2\right)....\left(-9\right)}{2.3.....10}.\dfrac{3.4....11}{2.3.....10}\)
\(A=\dfrac{-1}{10}.\dfrac{11}{2}=-\dfrac{11}{20}\)
a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
a, \(\left(2-\dfrac{3}{2}\right)\left(2-\dfrac{4}{3}\right)\left(2-\dfrac{5}{4}\right)\left(2-\dfrac{6}{5}\right)\)
\(=\left(\dfrac{4}{2}-\dfrac{3}{2}\right)\left(\dfrac{6}{3}-\dfrac{4}{3}\right)\left(\dfrac{8}{4}-\dfrac{5}{4}\right)\left(\dfrac{10}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
b. \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)\(=\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-\dfrac{2003.2002}{2002}\) = \(\dfrac{1+2003.2001-2003.2002}{2002}\) = \(\dfrac{1+\left(2003\left(2001-2002\right)\right)}{2002}\) = \(\dfrac{1+2003.\left(-1\right)}{2002}\) = \(\dfrac{1+\left(-2003\right)}{2002}\) = \(\dfrac{-2002}{2002}=-1\)
Chúc nguyễn hồng nhung học tốt
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