`a,2^300=(2^3)^100=8^100`
`3^200=(3^2)^100=9^100`
Vì `8<9` nên `2^300<3^200`
`b,E=[a+1]/[a-2]=[a-2+3]/[a-2]=1+3/[a-2]`
Để `E∈ZZ`
`⇒ 3/[a-2]∈ZZ`
`⇒ 3\vdotsa-2`
`⇒ a-2∈Ư(3)={1;3;-1;-3}`
Ta có bảng sau :
$\begin{array}{|l|r|} \hline a-2 &1&-1&3&-3\\ \hline a &3&2&5&-1\\ \hline \end{array}$
`d,` Ta có :
`([3x-5]/9)^2018+([3y+0,4]/3)^2020=0`
Vì $\left\{\begin{matrix}(\dfrac{3x-5}{9})^{2018}≥0 & \\(\dfrac{3y+0,4}{3})^{2020}≥0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}(\dfrac{3x-5}{9})^{2018}=0 & \\(\dfrac{3y+0,4}{3})^{2020}=0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}\dfrac{3x-5}{9}=0 & \\\dfrac{3y+0,4}{3}=0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}3x-5=0 & \\3y+0,4=0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}3x=5 & \\3y=-0,4& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}x=\dfrac{5}3 & \\y=\dfrac{-0,4}3& \end{matrix}\right.$
Vậy `(x;y)=(5/3;[-0,4]/3)`
Xin hay nhất !
