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a) Ta có:
\(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\) Q có nghĩa khi:
\(\left(1-3x\right)\left(x+\dfrac{1}{2}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x\ge0\\x+\dfrac{1}{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-3x\le0\\x+\dfrac{1}{2}\le\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le1\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\le x\le\dfrac{1}{3}\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{3}\)
b) Ta có: \(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\)
\(Q=\sqrt{x+\dfrac{1}{2}-3x^2-\dfrac{3}{2}x}\)
\(Q=\sqrt{-\left(3x^2+\dfrac{1}{2}x-\dfrac{1}{2}\right)}\)
\(Q=\sqrt{-3\left(x^2+\dfrac{1}{6}x-\dfrac{1}{6}\right)}\)
\(Q=\sqrt{-3\left(x^2+2\cdot\dfrac{1}{12}\cdot x+\dfrac{1}{144}-\dfrac{25}{144}\right)}\)
\(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\)
Mà: \(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\le\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)
Dấu "=" xảy ra khi:
\(\Leftrightarrow-3\left(x+\dfrac{1}{12}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{1}{12}=0\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
Vậy: \(Q_{max}=\dfrac{5}{12}.khi.x=-\dfrac{1}{12}\)
a) ĐKXĐ: thỏa mãn với mọi a thực
b) ĐKXĐ: \(\frac{1}{2a+1}>0\)
\(\Rightarrow2a+1>0\Rightarrow2a>-1\Leftrightarrow a>-\frac{1}{2}\)
c) ĐKXĐ: \(a\left(1-a\right)\ge0\)
+ Nếu: \(\hept{\begin{cases}a\ge0\\1-a\ge0\end{cases}}\Leftrightarrow1\ge a\ge0\)
+ Nếu: \(\hept{\begin{cases}a\le0\\1-a\le0\end{cases}\Rightarrow}\hept{\begin{cases}a\le0\\a\ge1\end{cases}}\)(vô lý)
Vậy \(0\le a\le1\)
d) ĐKXĐ: \(\frac{2}{\left(a-2\right)\left(a+3\right)}>0\)
\(\Rightarrow\left(a-2\right)\left(a+3\right)>0\)
+ Nếu: \(\hept{\begin{cases}a-2>0\\a+3>0\end{cases}}\Rightarrow a>2\)
+ Nếu: \(\hept{\begin{cases}a-2< 0\\a+3< 0\end{cases}}\Rightarrow a< -3\)
Vậy \(\orbr{\begin{cases}a>2\\a< -3\end{cases}}\)
Để biểu thức có nghĩa thì :
\(\sqrt{4+a^2}\left(đk:\forall a-tmđk\right)\)
\(\sqrt{\frac{1}{2a+1}}\left(đk:a\ne-\frac{1}{2};a\ge-\frac{1}{2}\Leftrightarrow a>-\frac{1}{2}\right)\)
\(\sqrt{a\left(1-a\right)}\left(đk:a\ge0\right)\)
\(\sqrt{\frac{2}{\left(a-2\right)\left(a+3\right)}}\left(đk:a\ge2;a\ne2\Leftrightarrow a>2\right)\)
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}=a\sqrt{\frac{1}{a+b}.\frac{1}{c+a}}\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}\)
Tương tự 2 cái còn lại cộng lại ta đc \(VT\le\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Cach khac
Dat \(P=\frac{a}{\sqrt{bc\left(1+a^2\right)}}+\frac{b}{\sqrt{ca\left(1+b^2\right)}}+\frac{c}{\sqrt{ab\left(1+c^2\right)}}\)
Ta co:
\(a+b+c=abc\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Dat \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(\Rightarrow xy+yz+zx=1\)
\(\Rightarrow P=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
Ta lai co:
\(\sqrt{\frac{yz}{1+x^2}}=\sqrt{\frac{yz}{xy+yz+zx+x^2}}=\sqrt{\frac{yz}{\left(x+y\right)\left(z+x\right)}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{z+x}\right)\)
Tuong tu:
\(\sqrt{\frac{zx}{1+y^2}}\le\frac{1}{2}\left(\frac{z}{y+z}+\frac{x}{x+y}\right)\)
\(\sqrt{\frac{xy}{1+z^2}}\le\frac{1}{2}\left(\frac{x}{z+x}+\frac{y}{y+z}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(\Rightarrow a=b=c=\sqrt{3}\)
Vay \(P_{min}=\frac{3}{2}\)khi \(a=b=c=\sqrt{3}\)
a)x thuộc R và x khác -1/2 và x khác 1/3