Question

Tìm giá trị nhỏ nhất của các biểu thức:
c) C=(x+2) mũ 20 +(y-1)mũ 20 +75
d) D=(x-1) ² + (x+y-5)mũ 4 -50
e) E=10/-x ²+3
Giúp mình nhanh với ạ huhu!

Answer 1

c: \(C=\left(x+2\right)^{20}+\left(y-1\right)^{20}+75>=75\)

Dấu '=' xảy ra khi x=-2 và y=1

d: \(D=\left(x-1\right)^2+\left(x+y-5\right)^4-50>=-50\)

Dấu '=' xảy ra khi x=1 và y=4

Answer 2

`c)`\(C=\left(x+2\right)^{20}+\left(y-1\right)^{20}+75\)

Ta có: \(\left\{{}\begin{matrix}\left(x+2\right)^{20}\ge0\\\left(y-1\right)^{20}\ge0\end{matrix}\right.\)

\(\Rightarrow C\ge75\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)

`d)`\(D=\left(x-1\right)^2+\left(x+y-5\right)^4-50\)

Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(x+y-5\right)^4\ge0\end{matrix}\right.\)

\(\Rightarrow D\ge-50\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

`e)`\(E=\dfrac{10}{-x^2+3}\)

Ta có: \(-x^2+3\le3\)

\(\Rightarrow E\ge\dfrac{10}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Answer 3

c) Có: \(\left\{{}\begin{matrix}\left(x+2\right)^{20}\ge0\\\left(y+1\right)^{20}\ge0\end{matrix}\right.\)

với mọi x,y

\(\Rightarrow\left(x+2\right)^{20}+\left(y+1\right)^{20}\ge0\)

\(\Rightarrow\left(x+2\right)^{20}+\left(y+1\right)^{20}+75\ge75\)

\(\Rightarrow C_{min}=75\)

Dấu ''='' xảy ra khi

\(\left\{{}\begin{matrix}\left(x+2\right)^{20}=0\\\left(y+1\right)^{20}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)

d) Có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(x+y-5\right)^4\ge0\end{matrix}\right.\)

với mọi x,y

\(\Rightarrow\left(x-1\right)^2+\left(x+y-5\right)^4\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(x+y-5\right)^4-50\ge-50\)

\(\Rightarrow D_{min}=-50\)

Dấu ''='' xảy ra khi

\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+y-5\right)^4=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)