Cho x , y là 2 số nguyên thỏa mãn` ( x – 2 )^2 + | x – 2y + 4 | = ( 1/2 + 1/3 + 1/4 + …. + 1/2021 ) . ( 2020 . 21212121 – 20202020 . 212121 )`
Cho x , y là 2 số nguyên thỏa mãn` ( x – 2 )^2 + | x – 2y + 4 | = ( 1/2 + 1/3 + 1/4 + …. + 1/2021 ) . ( 2020 . 21212121 – 20202020 . 212121 )`
Đáp án:
`x=2;y=3`
Giải thích các bước giải:
`( x – 2 )^2 + | x – 2y + 4 | = ( 1/2 + 1/3 + 1/4 + …. + 1/2021 ) . ( 2020 . 21212121 – 20202020 . 2121 )`
`=>( x – 2 )^2 + | x – 2y + 4 | = ( 1/2 + 1/3 + 1/4 + …. + 1/2021 ) . ( 2020 . 10001 . 2121- 20202020 . 2121)`
`=>( x – 2 )^2 + | x – 2y + 4 | = ( 1/2 + 1/3 + 1/4 + …. + 1/2021 ) . ( 20202020.2121 – 20202020 . 2121)`
`=>( x – 2 )^2 + | x – 2y + 4 | = ( 1/2 + 1/3 + 1/4 + …. + 1/2021 ) . 0`
`=>(x-2)^2+|x-2y+4|=0`
Ta có:
`(x-2)ge0` với mọi `x`
`|x-2y+4|ge0` với mọi `x`
Mà `(x-2)^2+|x-2y+4|=0`
`=>`$\left\{\begin{matrix}
(x-2)^2=0 & & \\
|x-2y+4|=0 & &
\end{matrix}\right.$
`=>`$\left\{\begin{matrix}
x-2=0 & & \\
x-2y+4=0 & &
\end{matrix}\right.$
`=>`$\left\{\begin{matrix}
x=2 & & \\
2-2y=-4 & &
\end{matrix}\right.$
`=>`$\left\{\begin{matrix}
x-2 & & \\
2y=6 & &
\end{matrix}\right.$
`=>`$\left\{\begin{matrix}
x=2 & & \\
y=3 & &
\end{matrix}\right.$
Vậy `x=2;y=3`
`(x-2)^2+//x-2y+4//=(1/2 +1/3 +1/4 +…+1/(2021)).(2020.21212121-20202020-2121)`
`⇔(x-2)^2+//x-2y+4//=(1/2 +1/3 +1/4 +…+1/(2021)).(2020.2121.10001-2020.100012121)`
`⇔(x-2)^2+//x-2y+4//=(1/2 +1/3 +1/4 +…+1/(2021)).0`
`⇔(x-2)^2+//x-2y+4//=0`
`⇒x-2=0 ` và `x-2y+4=0`
`⇒x=2` và `y=(x+4)/2 =(2+4)/2 =3`