ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne0\)

\(\dfrac{1}{x^2}-\dfrac{1}{x}=\sqrt{2x+1}-\sqrt{x+2}\)

\(\Leftrightarrow-\dfrac{x-1}{x^2}=\dfrac{x-1}{\sqrt{2x+1}+\sqrt{x+2}}\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}\right)=0\)

\(\Leftrightarrow x-1=0\) (do \(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}\) luôn dương)

\(\Leftrightarrow x=1\)

Đk: \(x\ge-\dfrac{1}{2},x\ne0\)

pt \(\Leftrightarrow\dfrac{1}{x^2}-\dfrac{1}{x}=\sqrt{2x+1}-\sqrt{x+2}\)

\(\Leftrightarrow\dfrac{1-x}{x^2}=\dfrac{2x+1-\left(x+2\right)}{\sqrt{2x+1}+\sqrt{x+2}}\)

\(\Leftrightarrow\dfrac{1-x}{x^2}=\dfrac{x-1}{\sqrt{2x+1}+\sqrt{x+2}}\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}\right)=0\)

\(\Leftrightarrow x=1\) (vì \(\dfrac{1}{\sqrt{2x+1}+\sqrt{x+2}}+\dfrac{1}{x^2}>0\))

Vậy \(S=\left\{1\right\}\)

-12/18 - (-21/35)

= -2/3 + 3/5

= -10/15 + 9/15

= -1/15

\(\dfrac{-12}{18}\) - \(\dfrac{-21}{35}\) 

=  \(\dfrac{-2}{3}\) + \(\dfrac{3}{5}\)

= \(\dfrac{-10}{15}\) + \(\dfrac{9}{15}\)

= \(-\dfrac{1}{15}\)

S =  \(\dfrac{1}{1.3}\)+\(\dfrac{1}{2.4}\)+...+\(\dfrac{1}{97.99}\)+\(\dfrac{1}{98.100}\) - \(\dfrac{49}{99}\)

S = (\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{97.99}\))+(\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+\(\dfrac{1}{6.8}\)+...+\(\dfrac{1}{98.100}\))- \(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+...+\(\dfrac{2}{97.99}\))+\(\dfrac{1}{2}\)(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))-\(\dfrac{49}{99}\)

S =\(\dfrac{1}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\))+\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))-\(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{99}\))+\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{100}\)) - \(\dfrac{49}{99}\)

S = \(\dfrac{1}{2}\).\(\dfrac{98}{99}\) + \(\dfrac{1}{2}\).\(\dfrac{49}{100}\) - \(\dfrac{49}{99}\)

S = \(\dfrac{49}{99}\) + \(\dfrac{49}{200}\) - \(\dfrac{49}{99}\)

S = (\(\dfrac{49}{99}\)- \(\dfrac{49}{99}\)) + \(\dfrac{99}{200}\)

S = 0 + \(\dfrac{49}{200}\)

S = \(\dfrac{49}{200}\)

\(8+2>8+1\)

>

Bài \(13\):
\(C=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{4-1}{1\cdot4}+\dfrac{9-4}{4\cdot9}+\dfrac{16-9}{9\cdot16}+...+\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)
\(=1-\dfrac{1}{\left(n+1\right)^2}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Bài \(10\):
\(B=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\cdot4}\)
\(=7\left(\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{28}\right)=7\cdot\dfrac{13}{28}=\dfrac{13}{4}\)

\(3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{2x}{4}=\dfrac{3y}{9}\) \(\left(1\right)\)
\(y=2z\Rightarrow\dfrac{3y}{3}=2z\Rightarrow\dfrac{3y}{9}=\dfrac{2z}{3}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau
ta có: \(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2z}{3}=\dfrac{2x+3y-2z}{4+9-3}=\dfrac{40}{10}=4\)
+\(\dfrac{2x}{4}=4\Rightarrow2x=16\Rightarrow x=8\)
+\(\dfrac{3y}{9}=4\Rightarrow3y=36\Rightarrow y=12\)
+\(\dfrac{2z}{3}=4\Rightarrow2z=12\Rightarrow z=6\)
Vậy \(x=8;y=12;z=6\)