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to cao luon

\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\\ =\dfrac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\dfrac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+....+\dfrac{1}{\sqrt{24.25}\left(\sqrt{25}+\sqrt{24}\right)}\\ =\dfrac{\sqrt{2}-1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}.\sqrt{3}}+...+\dfrac{\sqrt{25}-\sqrt{24}}{\sqrt{25}.\sqrt{24}}\\ =1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\\ =1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\)

=1-1/5=4/5

\(1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)

\(\Rightarrow\frac{31}{30}:\left(\frac{145}{6}-\frac{121}{5}\right)-\frac{\frac{3}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(\frac{-16}{15}\right):\left(\frac{41}{5}-\frac{25}{3}\right)\)

\(\Rightarrow\frac{31}{30}:\left(\frac{725}{30}-\frac{726}{30}\right)-\frac{\frac{6}{4}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(\frac{-16}{15}\right):\left(\frac{123}{15}-\frac{125}{15}\right)\)

\(\Rightarrow\frac{31}{30}:\frac{-1}{30}-\frac{\frac{3}{4}}{4x-\frac{1}{2}}=\frac{-16}{15}:\frac{-2}{15}\)

\(\Rightarrow\frac{31}{30}.\frac{30}{-1}-\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\frac{-16}{15}.\frac{15}{-2}\)

\(\Rightarrow\left(-31\right)-\frac{3}{4}:\left(4x-\frac{1}{2}\right)=8\)

\(\Rightarrow\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\left(-31\right)-8\)

\(\Rightarrow\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\left(-39\right)\)

\(\Rightarrow4x-\frac{1}{2}=\frac{3}{4}:\left(-39\right)\)

\(\Rightarrow4x-\frac{1}{2}=\frac{3}{4}.\frac{-1}{39}\)

\(\Rightarrow4x-\frac{1}{2}=\frac{-1}{52}\)

\(\Rightarrow4x=\frac{-1}{52}+\frac{1}{2}\)

\(\Rightarrow4x=\frac{-1}{52}+\frac{26}{52}\)

\(\Rightarrow4x=\frac{25}{52}\)

\(\Rightarrow x=\frac{25}{52}:4\)

\(\Rightarrow x=\frac{25}{52}.\frac{1}{4}\)

\(\Rightarrow x=\frac{25}{208}\)

Vậy \(x=\frac{25}{208}\)

Chúc bn học tốt

thank you

Đặt :

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

\(\Leftrightarrow2A=3+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}\)

\(\Leftrightarrow2A-A=\left(3+\dfrac{1}{2}+....+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\right)\)

\(\Leftrightarrow A=2-\dfrac{1}{2^{99}}\)

Vậy..

a: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

b: \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)

\(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1+1}\)

\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)

\(=\dfrac{2}{\sqrt{6}\left(\sqrt{6}+2\right)}=\dfrac{2}{6+2\sqrt{6}}=\dfrac{1}{3+\sqrt{6}}=\dfrac{3-\sqrt{6}}{3}\)

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{10}+1\right)+1\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{20}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{20}+1\right)+1\)

\(=2^{40}-1+1=2^{40}\)

\(1-\frac{2}{1.2.2.3:4}-\frac{3}{2.3.3.4:4}-...-\frac{25}{24.25.25.26:4}\)

\(=1-\left(\frac{4}{1.2.3}+\frac{4}{2.3.4}+\frac{4}{3.4.5}+...+\frac{4}{24.25.26}\right)\)

\(=1-2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{24.25}-\frac{1}{25.26}\right)\)

\(=1-2\left(\frac{1}{1.2}-\frac{1}{25.26}\right)\)

\(=\frac{1}{325}\)