Cho \(0< x< \frac{1}{2}\).Tìm Min \(P=\sqrt{x}+\sqrt[3]{1-2x}+x\)
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27 tháng 7 2018
Ukm
It's very hard
l can't do it
Sorry!
HT
12 tháng 6 2019
Bạn vt đề bài rõ ra nhé, mk RG trc rùi phần câu hỏi xem sau( P là j z?)
\(=\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}-2\)
\(=x-\sqrt{x}-3\)
14 tháng 3 2018
Đặt A là biểu thức cần CM
ví dụ Từ ĐK a + b + c = 3 => a² + b² + c² ≥ 3 ( Tự chứng minh )
Áp dụng BĐT quen thuộc x² + y² ≥ 2xy
a^4 + b² ≥ 2a²b (1)
b^4 + c² ≥ 2b²c (2)
c^4 + a² ≥ 2c²a (3)
28 tháng 3 2019
\(f\left(x;y\right)=x+y+x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)=\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{1}{2}\left(x\sqrt{3-3y^2}+y\sqrt{3-3x^2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{4}+x^2+\frac{3}{4}+y^2}{2}+\frac{1}{2}\left(\frac{-3x^2+y^2+3-3y^2+x^2+3}{2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{2}+x^2+y^2-x^2-y^2+3}{2}=\frac{9}{4}\)
\(\Rightarrow f\left(x;y\right)\le\frac{3\sqrt{3}}{2}\)
Dấu "=" khi x = y = \(\frac{\sqrt{3}}{2}\).
#Kaito#
28 tháng 8 2018
\(ĐKXĐ:\)tự làm nhé
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{-3\sqrt{x}-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(P=\left(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\right):\left(\frac{1+\sqrt{x}}{\sqrt{x}-3}\right)\)
\(P=\left(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\right)\times\left(\frac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(P=\frac{-3}{\sqrt{x}+3}\)
P/s tham khảo
5 tháng 4 2020
\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{x^2-2x+1}{2}\)
a)
Đkxđ:\(\left\{{}\begin{matrix}x-1\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
\(=\)\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}-2x}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(1+\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(x-1\right)}{2\left(\sqrt{x}+1\right)}=\frac{-2\sqrt{x}\left(x-1\right)}{2\sqrt{x}+2}\)