Question

[Linh Su Bông]

BT:Tìm x,y thoả mãn :

a) | x - 2017 | + | y- 2018 | < 0

b) 3.| x - y |⁵ + 10.| y + \(\dfrac{2}{3}\)|⁷ \(\le\) 0

c) \(\dfrac{1}{2}\). ( \(\dfrac{3}{4}x-\dfrac{1}{2}\))²⁰¹⁸ + \(\dfrac{2017}{2018}\).|\(\dfrac{4}{5}y+\dfrac{6}{25}\)| \(\le\)0

d) 2017.|2x - y|²⁰¹⁸ + 2018.|y - 4 | \(\le\) 0

... Helpppp Meeee vs các bn yêu dấu !!!!

Answer 1

[♥ Aoko ♥]

a) Ta có:

\(\left|x-2017\right|\ge0\) với \(\forall x\)

\(\left|y-2018\right|\ge0\) với \(\forall x\)

\(\Rightarrow\left|x-2017\right|+\left|y-2018\right|\ge0\) với \(\forall x\)

\(\Rightarrow\) Không có giá trị của x; y thỏa mãn yêu cầu

Vậy \(x;y\in\varnothing\)

b) Ta có:

\(3.\left|x-y\right|^5\ge0\)

\(10.\left|y+\dfrac{2}{3}\right|^7\ge0\)

\(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\ge0\left(1\right)\)

Theo bài ra ta có: \(3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7\le0\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow3.\left|x-y\right|^5+10.\left|y+\dfrac{2}{3}\right|^7=0\)

\(\Rightarrow\left\{{}\begin{matrix}3.\left|x-y\right|^5=0\\10.\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|^5=0\\\left|y+\dfrac{2}{3}\right|^7=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x-y=0\\y+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=y\\y=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{-2}{3}\end{matrix}\right.\)\(\)