chứng minh rằng: (\(8^{10}-8^9-8^8\)) : 55 là số tự nhiên
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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
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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-\(\sqrt{11}\)
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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 \(\left(3x-5\right)^{2008}+\left(y^2-1\right)^{2010}+\left(x-z\right)^{2012}=0\)
\(\Rightarrow\hept{\begin{cases}\left(3x-5\right)^{2008}=0\\\left(y^2-1\right)^{2010}=0\\\left(x-z\right)^{2012}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1;-1\\z=x=\frac{5}{3}\end{cases}}\)
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810 - 89 - 88 : 55
ta có 810 - 89 - 88
= 88( 8^2 - 8 - 1)
=88 . ( 64 - 8 - 1)
= 88 .55 luôn chia hết cho 55
810 - 89 - 88 : 55 là số tự nhiên
thanks pn nhìu na Lê Mạnh Tiến Đạt