a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

a/ ĐKXĐ: \(x\ne\left\{8;9;10;11\right\}\)

\(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

\(\Leftrightarrow\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)

\(\Leftrightarrow x\left(\frac{1}{x-8}-\frac{1}{x-9}+\frac{1}{x-11}-\frac{1}{x-10}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-9}-\frac{1}{x-8}=\frac{1}{x-11}-\frac{1}{x-10}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{\left(x-9\right)\left(x-8\right)}=\frac{1}{\left(x-11\right)\left(x-10\right)}\)

\(\Leftrightarrow x^2-17x+72=x^2-21x+110\)

\(\Rightarrow x=\frac{19}{2}\)

b/ ĐK: \(x\ne\left\{3;4;5;6\right\}\)

\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}-\frac{1}{x-5}=\frac{1}{x-4}-\frac{1}{x-6}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\frac{-2}{\left(x-3\right)\left(x-5\right)}=\frac{-2}{\left(x-4\right)\left(x-6\right)}\)

\(\Leftrightarrow x^2-8x+15=x^2-10x+24\)

\(\Rightarrow x=\frac{9}{2}\)

Theo mình thì VP = -5, vì nếu VP = 0 thì theo cách làm của mình x = \(\frac{-531}{23}\) (số to quá nên mình nghĩ là đề sai)

\(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=5\)

\(\Leftrightarrow\) \(\frac{x+1}{4}-1-\frac{x+2}{5}-1+\frac{x+5}{8}-1+\frac{x+7}{10}-1-\frac{x+9}{12}-1=0\)

\(\Leftrightarrow\) \(\frac{x-3}{4}-\frac{x-3}{5}+\frac{x-3}{8}+\frac{x-3}{10}-\frac{x-3}{12}=0\)

\(\Leftrightarrow\) (x - 3)(\(\frac{1}{4}-\frac{1}{5}+\frac{1}{8}+\frac{1}{10}-\frac{1}{12}\)) = 0

\(\Leftrightarrow\) x - 3 = 0

\(\Leftrightarrow\) x = 3

Vậy S = {3}

Chúc bn học tốt!!

= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)

= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)

= \(x^8.\frac{1}{10}.1.1.1.1\)

= \(x^8.\frac{1}{10}\)

Mk ko pik co dung ko nua

=4/10

\(\frac{x^8}{10}\)

\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)

\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)

\(< =>4x-12-4x+2=10x+10+5\)

\(< =>10x=-10-10-5=-25\)

\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)

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

\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)

\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)

\(\frac{x-1}{4}-\frac{5-2x}{9}=3x-\frac{2}{3}\)

\(< =>\frac{\left(x-1\right).9}{36}-\frac{\left(5-2x\right).4}{36}=\frac{3x.36}{36}-\frac{2.12}{36}\)

\(< =>\left(x-1\right).9-\left(5-2x\right).4=108x-24\)

\(< =>9x-9-20+8x=108x-24\)

\(< =>108x-17x=-29+24\)

\(< =>91x=-5< =>x=-\frac{5}{91}\)

\(\frac{x+1}{10}+\frac{x+2}{9}+\frac{x+3}{8}+\frac{x+4}{7}+\frac{x+5}{6}=-5\)

\(\left(\frac{x+1}{10}+1\right)+\left(\frac{x+2}{9}+1\right)+\left(\frac{x+3}{8}+1\right)+\left(\frac{x+4}{7}+1\right)+\left(\frac{x+5}{6}+1\right)=0\)

\(\frac{x+11}{10}+\frac{x+11}{9}+\frac{x+11}{8}+\frac{x+11}{7}+\frac{x+12}{6}=0\)

\(\left(x+11\right)\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}\right)=0\)

Vì : \(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}>0\)

\(\Rightarrow x+11=0\)

\(\Rightarrow x=-11\)