Ta thấy :

\(\frac{20}{39}>\frac{14}{39}\)

\(\frac{22}{27}>\frac{22}{29}\)

\(\frac{18}{43}< \frac{18}{41}\)

\(\Rightarrow\frac{20}{39}+\frac{22}{27}+\frac{18}{43}>\frac{14}{39}+\frac{22}{29}+\frac{18}{41}\)

=> A > B

X=-1

mk chỉ biết kết quả thôi

bạn ns vậy thì ai mà chả bt! 

Ta có : 

\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)

\(\Rightarrow\frac{1}{243^9}>\frac{1}{8^{13}}\)

| x - 2,2 | = | 0,2 - x |

\(\Rightarrow\orbr{\begin{cases}x-2,2=0,2-x\\x-2,2=-\left(0,2-x\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+x=0,2+2,2\\x-2,2=-0,2+x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=2,4\\x-x=-0,2+2,2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1,2\\0=2\left(\text{loại}\right)\end{cases}}\)

bạn chắc không

(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0  
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0  
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0  
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0  
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Leftrightarrow\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+349-20}{5}=0\)

\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)

\(\Rightarrow x+329=0\)\(\Leftrightarrow x=-329\)

Vậy \(x=-329\)