cho x,y,z >0 thỏa mãn \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1\)
khi đó giá trị nhỏ nhất của A=\(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)
lÀ .....
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Vì \(\left(x-2\right)^4\ge0;\left(2y-1\right)^{2004}\ge0\forall x;y\)
\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2004}\ge0\forall x;y\)
Mà đề lại cho \(\left(x-2\right)^4+\left(2y-1\right)^{2004}\le0\Rightarrow\left(x-2\right)^4=0;\left(2y-1\right)^{2004}=0\)
\(\Rightarrow x=2;y=\frac{1}{2}\) Thay vào đa thức \(21x^{2y}+4xy^2\) ta được :
\(21.2^{2.\frac{1}{2}}+4.2.\left(\frac{1}{2}\right)^2=21.2+8.\frac{1}{4}=42+2=44\)
là 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
áp dụng bđt Schwarz thôi mak :
A >/ (x+y+z)/2
phần còn lại là c/m x+y+z >/ căn xy + căn yz + căn zx >/ 1 =>A >/ 1/2
thật lòng xin lỗi anh chị , em mới hok lớp 6 hà !!!!!!