\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)

\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1}{2009}\)

1,

\(| x - \frac{2}{7} | = \frac{-1}{5}.\frac{-5}{7}\)

\(|x- \frac{2}{7}|=\frac{1}{7}\)

<=> \(x- \frac{2}{7} = \frac{1}{7} => x= \frac{3}{7} \)

Và \(x - \frac{2}{7} =\frac{-1}{7} => x= \frac{1}{7}\)

Học tốt

\(5^{61}+25^{31}+125^{21}\)

\(=5^{61}+\left(5^2\right)^{31}+\left(5^3\right)^{21}\)

\(=5^{61}\cdot5^{2\cdot31}\cdot5^{3\cdot21}\)

\(=5^{61}+5^{62}+5^{63}\)

\(=5^{61}\cdot\left(1+5+5^2\right)\)

\(=5^{61}\cdot\left(6+5^2\right)\)

\(=5^{61}\cdot\left(6+25\right)\)

\(=5^{61}\cdot31\)

Vì \(5^{61}\inℤ\)

\(\Rightarrow5^{61}\cdot31⋮31\)

\(\Rightarrow5^{61}+25^{31}+125^{21}⋮31\)

Vậy bài toán đã được chứng minh . 

\(A=1+3+3^2+3^3+.....+3^{31}\)

\(\Rightarrow3A=3+3^2+3^3+.....+3^{32}\)

\(\Rightarrow3A-A=2A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{2}\left(đpcm\right)\)

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

Đặt A là biểu thức trên

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{31}{15^2.16^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{31}{225.256}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{225}-\frac{1}{256}\)

\(=1-\frac{1}{256}=\frac{255}{256}< 1\)

Vậy...