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

=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

=>\(\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x}=\frac{1}{2011}\)

=>\(\frac{1}{x}-\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2011}\)

=>\(0-\frac{1}{x+1}=\frac{1}{2011}\)

=>\(-\frac{1}{x+1}=\frac{1}{2011}\)

=>-x+1=2011

=>-x=2011-1

=>-x=2010

=>x=-2010

Vậy x=-2010

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

<=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

<=>\(-\frac{1}{x+1}=\frac{1}{2011}\)

<=>-x-1=2011

<=>x=-2012

Đáp số: \(x=-2012\)

<=>\(\frac{1}{x}-\frac{1}{x+1}\)=\(\frac{1}{x+}+\frac{1}{2011}\)

<=>\(\frac{1}{x}+\left(\frac{-1}{x+1}\right)=\frac{1}{x}+\frac{1}{2011}\)

<=>\(-\frac{1}{x}=\frac{1}{2011}\)

<=>2011(-1)=x.1

<=>-2011=x

<=>x=-2011

vậy x=-2011

Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))

Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) . 

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0

Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)

\(\pm3\)

\(\frac{|x|}{186}=\left(1-\frac{30}{31}\right)+\left(\frac{60}{61}-1\right)\)

\(\Leftrightarrow|x|=186\left(\frac{1}{31}-\frac{1}{61}\right)\)

\(\Leftrightarrow|x|=6-\frac{186}{61}\)

\(\Leftrightarrow|x|=\frac{180}{61}\)

\(\Leftrightarrow x=\pm\frac{180}{61}\)

\(\frac{|x|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)\)

\(\frac{|x|}{186}=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)\)

\(\frac{|x|}{186}=1-\frac{30}{31}+\frac{61}{62}-1\)

\(\frac{|x|}{186}=\left(1-1\right)+\left(\frac{61}{62}-\frac{30}{31}\right)\)

\(\frac{|x|}{186}=\frac{1}{62}\)

\(\Rightarrow|x|=186.\frac{1}{62}\)

\(|x|=3\)

\(\Rightarrow x=\pm3\)

Thi vòng 12 à bạn!!! Để mk chép đề mà làm 

\(\left(\frac{1}{2}\right)^{x+4}=\left(\frac{1}{4}\right)^{\frac{3}{2}x-4}\)

=> \(\left(\frac{1}{2}\right)^{x+4}=\left(\frac{1}{2}\right)^{2\left(\frac{3}{2}x-4\right)}\)

=> \(\left(\frac{1}{2}\right)^{x+4}=\left(\frac{1}{2}\right)^{3x-8}\)

=> \(x+4=3x-8\)

=> \(3x-8-x=4\)

=> \(2x-8=4\)

=> \(2x=12\)

=> \(x=\frac{12}{2}=6\)

\(\left(\frac{1}{2}\right)^{-x+4}=\left(\frac{1}{4}\right)^{\frac{3}{2}x-4}\)

=>\(\left(\frac{1}{2}\right)^{-x+4}=\left(\frac{1}{2}\right)^{3x-8}\)

=>-x+4=3x-8

<=>4x=12

<=>x=3

Vậy x=3

\(\left(\frac{1}{4}\right)^{\frac{3}{2}-4}=\left(\frac{1}{2}\right)^{2.\left(\frac{3}{2}-4\right)}=\left(\frac{1}{2}\right)^{-1}\)

; do đó -x + 4 = -1

=> -x = -1 - 4 = -5

=> x = 5

<=> \(\left(\frac{1}{2}\right)^{-x+4}=\left(\frac{1}{2}\right)^{3x-4}< =>-x+4=3x-4< =>x=2\)