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\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)
\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1}{2009}\)
Ta có : \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{31}{15^2.16^2}\)
= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+...+\dfrac{16^2-15^2}{15^2.16^2}\)
= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{15^2}-\dfrac{1}{16^2}\)
= \(1-\dfrac{1}{16^2}< 1\)
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^2}+...+\frac{99}{3^{89}}-\frac{100}{3^{99}}\)
\(\Rightarrow4C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\left(1\right)\)
Đặt: \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=2+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow4C< B< \frac{3}{4}\)
\(\Rightarrow C< \frac{3}{16}\left(đpcm\right)\)
(Đánh nhanh quá sai chỗ nào thông cảm nha :))
Làm được mỗi câu a :)
\(\frac{x-3}{2}+\frac{x-3}{3}=\frac{x-3}{4}\)
\(\Leftrightarrow\frac{x-3}{2}+\frac{x-3}{3}-\frac{x-3}{4}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)=0\)
Vì \(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\ne0\) nên x - 3 = 0
Vậy x = 3
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
\(Q=1+3+3^2+3^3+...+3^{31}\)(có 32 số hạng)
\(3Q=3+3^2+3^3+3^4+...+3^{32}\)
\(3Q-Q=\left(3+3^2+3^3+3^4+...+3^{31}+3^{32}\right)-\left(1+3+3^2+3^3+...+3^{31}\right)\)
\(2Q=3^{32}-1\)
\(Q=\frac{3^{32}-1}{2}\)(đpcm)
sai