Bài làm

x = \(\frac{20}{21}+\frac{21}{22}+\frac{22}{23}+\frac{23}{20}\)

x = 1 + 1 + 1 + 1 + \((\)\(\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23})\)

Ta thấy 0 < \(\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\)

\(\Rightarrow\) 1 + 1 + 1 + 1 + \((\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23})\)> 4

\(\Rightarrow\)x > 4

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)

  \(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)

  \(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)

 \(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)

 \(x+\frac{8}{45}=-\frac{37}{45}\)

\(x=-\frac{37}{45}-\frac{8}{45}\)

\(x=-1\)

thế phần b, c đâu bạn

\(\left(\frac{x}{20}+1\right)+\left(\frac{x-1}{21}+1\right)=\left(\frac{x-2}{22}+1\right)+\left(\frac{x-3}{23}+1\right)\)

\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)

\(\left(x+20\right).\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)

mà \(\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)\ne0\)

=> x+20=0 => x=-20

vậy x=-20

\(\frac{x}{20}+\frac{x-1}{21}=\frac{x-2}{22}+\frac{x-3}{23}\)

\(1+\frac{x}{20}+1+\frac{x-1}{21}=1+\frac{x-2}{22}+1+\frac{x-3}{23}\)

\(\frac{x+20}{20}+\frac{21+x-1}{21}=\frac{22+x-2}{22}+\frac{23+x-3}{23}\)

\(\frac{x+20}{20}+\frac{x+20}{21}=\frac{x+20}{22}+\frac{x+20}{23}\)

\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)

\(\left(x+20\right)\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)

Mà \(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\ne0\)

\(\Rightarrow x+20=0\)

\(\Rightarrow x=-20\)

Vậy x = -20

\(e,\frac{22}{15}-x=-\frac{8}{27}\)

=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)

=> \(x=\frac{22}{15}+\frac{8}{27}\)

=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)

\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)

=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)

=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)

=> \(\frac{2x-5}{5}=-\frac{5}{4}\)

=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)

=> \(2x=-\frac{5}{4}\)

=> \(x=-\frac{5}{8}\)

\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)

=> \(-\frac{9}{4}x+\frac{37}{4}=20\)

=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)

=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)

\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)

=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)

=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)

=> \(-10\le x\le-\frac{13}{7}\)

Đến đây tìm x

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

\(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\)

\(-5x+9=6x-90\)

\(-5x-6x=-90-9\)

\(-11x=-99\)

\(x=\frac{-99}{-11}=9\)

b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)

\(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)

\(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)

\(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)

x=30

Chúc bạn học tốt!!

-Xét \(x\ge y\ge z\). Dễ cm bđt đúng

-Xét \(x\ge z\ge y\)

Đặt x=z+a, z=y+b với \(a,b\ge0\)

=>x=y+a+b

BĐT\(< =>\frac{x-y}{y\left(y+1\right)}\ge\frac{x-z}{x\left(x+1\right)}+\frac{z-x}{z\left(z+1\right)}\)

<=>\(\frac{a+b}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)}+\frac{b}{z\left(z+1\right)}\)

Vì \(x\ge z\ge y=>x\left(x+1\right)\ge z\left(z+1\right)\ge y\left(y+1\right)\)

\(=>\frac{a}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)},\frac{b}{y\left(y+1\right)}\ge\frac{b}{z\left(z+1\right)}\)

=>\(\frac{a+b}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)}+\frac{b}{z\left(z+1\right)}\)=>bđt cần cm đúng=>đpcm