tra Google

x=1/2,y=-2;z=1

Vậy 4x+y-3z=4.1/2+(-2)-3.1=-3

Ta có (2x-1)\(^{2016}\)+(3y+6)\(^{2014}\)+(z-1)\(^{2012}\)=0

\(\Leftrightarrow\)(2x-1)\(^{2016}\)=0 ; (3y+6)\(^{2014}\)=0 ; (z-1)\(^{2012}\)=0

Ta co :(2x-1)\(^{2016}\)=0\(\Rightarrow\)2x-1=0\(\Rightarrow\)2x=1\(\Rightarrow\)x=\(\frac{1}{2}\)

          (3y+6)\(^{2014}\)=0 \(\Rightarrow\)3y+6=0 \(\Rightarrow\)3y=-6 \(\Rightarrow\)y=-2

          (z-1)\(^{2012}\)=0 \(\Rightarrow\)z-1=0 \(\Rightarrow\)z=1

Vậy  4x+y-3z=4*\(\frac{1}{2}\)+(-2)-3*1=2-2-3=-3

ta có: N=6-\(\sqrt{11}\)

=\(\sqrt{36}\)-\(\sqrt{11}\)

ta có \(\sqrt{31}\)<\(\sqrt{36}\);\(\sqrt{13}\)>\(\sqrt{11}\)

\(\Rightarrow\)\(\sqrt{31}\)-\(\sqrt{13}\)<\(\sqrt{36}\)-\(\sqrt{11}\)

\(\Leftrightarrow\)\(\sqrt{31}\)-\(\sqrt{13}\)<6-\(\sqrt{11}\)

thanks pn nhìu na 

Ta có \(\hept{\begin{cases}\left(3x-5\right)^{2008}\ge0\\\left(y^2-1\right)^{2010}\ge0\\\left(x-z\right)^{2012}\ge0\end{cases}}\)mà \(\left(3x-5\right)^{2008}+\left(y^2-1\right)^{2010}+\left(x-z\right)^{2012}=0\)

\(\Rightarrow\hept{\begin{cases}\left(3x-5\right)^{2008}=0\\\left(y^2-1\right)^{2010}=0\\\left(x-z\right)^{2012}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1;-1\\z=x=\frac{5}{3}\end{cases}}\)

\(x^3-3x=0\\ \Leftrightarrow x\left(x^2-3\right)=0\\ \orbr{\begin{cases}x=0\\x^2-3=0\end{cases}}\\ \orbr{\begin{cases}x=0\\x^2=3\end{cases}}\\ \orbr{\begin{cases}x=0\\x=+_-\sqrt{3}\end{cases}}\)

x³ - 3x = 0

x.x² - 3x = 0

x.(x² - 3) = 0

=> x = 0 hoặc x² - 3 = 0

=> x = 0 hoặc x² = 3

=> x = 0 hoặc x = \(\sqrt{3}\)