CMR \(2x^3-3x^2-x+1< 0\)
Với x thuộc R thỏa \(\frac{1}{2}< x< \frac{1+\sqrt{5}}{2}\)
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Bạn vt đề bài rõ ra nhé, mk RG trc rùi phần câu hỏi xem sau( P là j z?)
\(=\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}-2\)
\(=x-\sqrt{x}-3\)
\(M=2\sqrt{3^2.3}-6\frac{\sqrt{2^2.3}}{3}+\frac{3}{5}\sqrt{5^2.3}\)
\(M=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{x-1}.\frac{\left|x-1\right|}{2x}=\frac{-2\left(x-1\right)}{\left(x-1\right).2x}=-\frac{1}{x}\)
\(f\left(x;y\right)=x+y+x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)=\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{1}{2}\left(x\sqrt{3-3y^2}+y\sqrt{3-3x^2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{4}+x^2+\frac{3}{4}+y^2}{2}+\frac{1}{2}\left(\frac{-3x^2+y^2+3-3y^2+x^2+3}{2}\right)\)
\(\Rightarrow\frac{\sqrt{3}}{2}f\left(x;y\right)\le\frac{\frac{3}{2}+x^2+y^2-x^2-y^2+3}{2}=\frac{9}{4}\)
\(\Rightarrow f\left(x;y\right)\le\frac{3\sqrt{3}}{2}\)
Dấu "=" khi x = y = \(\frac{\sqrt{3}}{2}\).
#Kaito#
\(E=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}\)
\(E=\frac{2}{x-1}\cdot\frac{\sqrt{\left(x-1\right)^2}}{\sqrt{\left(2x\right)^2}}\)
\(E=\frac{2}{x-1}\cdot\frac{-\left(x-1\right)}{2x}\)
\(E=\frac{-1}{x}\)
_________
\(G=\frac{x-16}{\sqrt{x-7}-3}\)
\(G=\frac{\left(\sqrt{x-7}-3\right)\left(\sqrt{x-7}+3\right)}{\sqrt{x-7}-3}\)
\(G=\sqrt{x-7}+3\)
\(2x^3-3x^2-x+1< 0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-x-1\right)< 0\)
Để bât đẳng thưc đung thì
\(\hept{\begin{cases}2x-1>0\\x^2-x-1< 0\end{cases}}\)hoặc \(\hept{\begin{cases}2x-1< 0\\x^2-x-1>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>\frac{1}{2}\\\frac{1-\sqrt{5}}{2}< x< \frac{1+\sqrt{5}}{2}\end{cases}}\)hoặc \(\Leftrightarrow\hept{\begin{cases}x< \frac{1}{2}\\\frac{1-\sqrt{5}}{2}>x;x>\frac{1+\sqrt{5}}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}< x< \frac{1+\sqrt{5}}{2}\\x< \frac{1-\sqrt{5}}{2}\end{cases}}\)
Vậy ta co ĐPCM