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

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

Bác viết nhộn đề gồi :v

\(.\frac{x+4}{20}+\frac{x+3}{21}+\frac{x+2}{22}+\frac{x+1}{23}=-4\)

\(\Rightarrow\frac{x+4}{20}+1+\frac{x+3}{21}+1+\frac{x+2}{22}+1+\frac{x+1}{23}+1=0\)

\(\Rightarrow\frac{x+24}{20}+\frac{x+24}{21}+\frac{x+24}{22}+\frac{x+24}{23}=0\)

\(\Rightarrow\left(x+24\right)\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)

=> x=-24

    \(\frac{x+4}{20}+\frac{x+3}{21}\frac{x+2}{22}+\frac{x+1}{23}\)\(=-4\)

\(\Rightarrow\left(\frac{x+4}{20}+1\right)+\left(\frac{x+3}{21}+1\right)+\left(\frac{x+2}{22}+1\right)\)\(+\left(\frac{x+1}{23}+1\right)=0\)

\(\Rightarrow\left(\frac{x+4}{20}+\frac{20}{20}\right)+\left(\frac{x+3}{21}+\frac{21}{21}\right)\)\(+\left(\frac{x+2}{22}+\frac{22}{22}\right)+\left(\frac{x+1}{23}+\frac{23}{23}\right)=0\)

\(\frac{\Rightarrow x+24}{20}+\frac{x+24}{21}+\frac{x+24}{22}+\frac{x+24}{23}=0\)

\(\Rightarrow\left(x+24\right)+\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)

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

\(\Rightarrow x+24=0\)

\(\Rightarrow x=24\)

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

Hnay đi học, cô giáo có sửa cho bạn bài đó hong dọ, do cô mình giao cái bài về nhà  y sì dãy í, mà mai nộp ròi, nhưng mình k biết làm, nếu bạn biết , chỉ mình với :((

Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)

\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)

\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)thỏa mãn đề.