Question

tìm x,y biết

(1/2x - 5)^20+(y^2 - 1/4)^10 < 0

Answer 1

ta có \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\end{cases}\Rightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}}\)

Answer 2

Bài làm:

Ta có: \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\left(\forall x\right)\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\left(\forall y\right)\end{cases}\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\left(\forall x,y\right)}\)

Mà theo đề bài: \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)

\(\Rightarrow\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

Answer 3

Vì \(\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\);\(\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\)

\(\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\)

Mà theo đề bài thì \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)

Vậy tồn tại một trường hợp duy nhất là \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=10\\y=\frac{1}{2}\end{cases}}\)

Answer 4

\(y=\pm\frac{1}{2}\) nhé:v

Answer 5

Ta có \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall x;y\)

Dấu"=" xảy ra <=> \(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}}\Rightarrow\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

Vậy các cặp (x;y) thỏa mãn là \(\left(10;\frac{1}{2}\right);\left(10;-\frac{1}{2}\right)\)

Answer 6

\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\) (*)

Ta có : \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{1}{2}x-5\right)+\left(y^2-\frac{1}{4}\right)\ge0\forall x,y\)

Kết hợp với (*)

=> (*) chỉ xảy ra trường hợp \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)=0\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

=> \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)<=> \(\left(x;y\right)=\left\{\left(10;\frac{1}{2}\right),\left(10;-\frac{1}{2}\right)\right\}\)

Answer 7

cảm ơn mọi người nhìu