Giải thích các bước giải:
Bài 5:
a) Có:
`7^(n+3)-7^(n+2)+7^(n+1)-7^n`
`=7^(n+1).(7^2-7)+7^(n-1).(7^2-7)`
`=7^(n+1).42+7^(n-1).42`
`=42.(7^(n+1)+7^(n-1))⋮42` (vì `42⋮42`).
Vậy `7^(n+3)-7^(n+2)+7^(n+1)-7^n⋮42.`
b) Có:
`3^n+3^(n+1)+3^(n+2)+3^(n+3)+3^(n+4)+3^(n+5)`
`=3^(n-1).(3+3^2+3^3)+3^(n+2).(3+3^2+3^3)`
`=3^(n-1).39+3^(n+2).39`
`=39.(3^(n-1)+3^(n+2))⋮39` (vì `39⋮39`)
Vậy `3^n+3^(n+1)+3^(n+2)+3^(n+3)+3^(n+4)+3^(n+5)⋮39.`
Bài 6:
a) Ta thấy:
`(3x-5)^2018>=0`
`(y^2-1)^2020>=0`
`(x-z)^2010>=0`
`=>(3x-5)^2018+(y^2-1)^2020+(x-z)^2010>=0`
Dấu $"="$ xảy ra $\Leftrightarrow \left\{\begin{matrix} 3x-5=0 \\ y^2-1=0 \\ x-z=0 \end{matrix}\right.$
$\Rightarrow \left\{\begin{matrix} 3x=5 \\ y^2=1 \\ x=z \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\dfrac{5}{3} \\ y=±1 \\ z=\dfrac{5}{3} \end{matrix}\right.$
Vậy $x=\dfrac{5}{3};y=±1;z=\dfrac{5}{3}.$
b) Ta thấy:
`(x-2)^4>=0`
`(2y-1)^2018>=0`
`=>(x-2)^4+(2y-1)^2018>=0`
Mà `(x-2)^4+(2y-1)^2018<=0`
`=>(x-2)^4+(2y-1)^2018=0`
$\Rightarrow \left\{\begin{matrix} x-2=0 \\ 2y-1=0 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=2 \\ 2y=1 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=2 \\ y=\dfrac{1}{2} \end{matrix}\right.$
Vậy $x=2;y=\dfrac{1}{2}.$
