Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

           \(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

             \(.\)                   \(.\)

             \(.\)

             \(.\)                    \(.\)  

             \(.\)                    \(.\)

         \(\frac{1}{2013^2}< \frac{1}{2012\cdot2013}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{2013^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}\)

Mà \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}=1-\frac{1}{2013}< 1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2013^2}< 1\)

Nhớ k cho mình nhé!

Chúc các bạn học tốt!

mình giải ở đè trước rồi

a, 11 1/4-(2 5/7+5 1/4)

= 45/4-(19/7+21/4)

= 45/4-223/28

=23/7

b, (8 5/11+3 5/8)-3 5/11

=(93/11+29/8)-38/11

=1063/88-38/11

=69/8

a,  =\(11\frac{1}{4}-2\frac{5}{7}-5\frac{1}{4}\)

\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)

\(=6-2\frac{5}{7}\)

\(=\frac{23}{7}\)

b,  \(=8\frac{5}{11}+3\frac{5}{8}-3\frac{5}{11}\)

\(=\left(8\frac{5}{11}-3\frac{5}{11}\right)+3\frac{5}{8}\)

\(=5+3\frac{5}{8}\)

\(=\frac{69}{8}\)

Bạn Pham Ngoc Anh ơi ! Sorry bạn nhé ! mik chỉ có thể k 1 lần thay vì không k được mik sẽ kết bạn nhé !

Đặt x=a-2,ta có : \(P=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\left(\frac{3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{2\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\frac{3}{3-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)

\(=\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)

a/b=8

Ai biết cách làm, làm ơn ghi rõ ra dùm mik nhe. Cảm ơn nhiều trước.

\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)

\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)

\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)

\(\left(4\frac{46}{65}+x\right).1\frac{1}{12}=5,75\)

\(\Rightarrow\frac{306}{65}+x.\frac{13}{12}=\frac{23}{4}\)

\(\Rightarrow\frac{51}{10}+\frac{13}{12}x=\frac{23}{4}\)

\(\Rightarrow306x=65x=345\)

\(\Rightarrow65x=39\)

\(\Rightarrow x=\frac{3}{5}\)

b, \(\frac{5}{4}-\left(\frac{3}{2}x+0,5\right)=1\frac{1}{4}\)

\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-0,5=\frac{5}{4}\)

\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-\frac{1}{2}=\frac{5}{4}\)

\(\Rightarrow\frac{3}{4}-\frac{3}{2}x=\frac{5}{4}\)

\(\Rightarrow3-6x=5\)

\(\Rightarrow-6x=2\)

\(\Rightarrow x=-\frac{1}{3}\)

\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)

\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)

\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)

\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)

\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)

\(A=1,3\left(2\right)\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

\(A=\frac{2^{100}-1}{2^{100}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

hok tốt!!

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Vậy \(A=1-\frac{1}{2^{100}}\)