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a) Ta có: \(2x^2+2x+3=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{5}{2}\)
\(=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)
\(\Rightarrow S\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)
Vậy \(S_{max}=\frac{6}{5}\Leftrightarrow\sqrt{2}x+\frac{1}{\sqrt{2}}=0\Leftrightarrow x=-\frac{1}{2}\)
b) Ta có: \(3x^2+4x+15=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\frac{2}{\sqrt{3}}+\frac{4}{3}+\frac{41}{3}\)
\(=\left(\sqrt{3}x+\frac{2}{\sqrt{3}}\right)^2+\frac{41}{3}\ge\frac{41}{3}\)
\(\Rightarrow T\le\frac{5}{\frac{41}{3}}=\frac{15}{41}\)
Vậy \(T_{max}=\frac{15}{41}\Leftrightarrow\sqrt{3}x+\frac{2}{\sqrt{3}}=0\Leftrightarrow x=\frac{-2}{3}\)
c) Ta có: \(-x^2+2x-2=-\left(x^2-2x+1\right)-1\)
\(=-\left(x-1\right)^2-1\le-1\)
\(\Rightarrow V\ge\frac{1}{-1}=-1\)
Vậy \(V_{min}=-1\Leftrightarrow x-1=0\Leftrightarrow x=1\)
d) Ta có: \(-4x^2+8x-5=-\left(4x^2-8x+5\right)\)
\(=-\left(4x^2-8x+4\right)-1\)
\(=-\left(2x-2\right)^2-1\le-1\)
\(\Rightarrow X\ge\frac{2}{-1}=-2\)
Vậy \(X_{min}=-2\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
a) A có nghĩa khi \(\hept{2x-2\ne02-2x^2\ne0\Leftrightarrow\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}\Leftrightarrow}x\ne\pm1}\)
Vậy A có nghĩa khi \(x\ne\pm1\)
b) \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\left(x\ne\pm1\right)\)
\(\Leftrightarrow A=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)
Vậy A=\(\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)
b) \(A=\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)
A=\(\frac{-1}{2}\)\(\Leftrightarrow\frac{1}{2\left(x-1\right)}=\frac{-1}{2}\)
\(\Leftrightarrow-2\left(x-1\right)=2\)
<=> x-1=-1
<=> x=0 (tmđk)
Vậy x=0 thì \(A=\frac{-1}{2}\)
\(B=\left(\frac{2x+1}{2x-1}+\frac{4}{1-4x^2}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\left(x\ne\pm\frac{1}{2}\right)\)
\(\Leftrightarrow B=\left(\frac{2x+1}{2x-1}-\frac{4}{4x^2-1}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(\Leftrightarrow B=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{4}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right)\cdot\frac{2x+1}{x^2+2}\)
\(\Leftrightarrow B=\frac{\left(2x\right)^2+2\cdot1\cdot2x+1-4-\left[\left(2x\right)^2-2\cdot2x\cdot1+1^2\right]}{\left(2x-1\right)\left(2x+1\right)}\cdot\frac{2x+1}{x^2+2}\)
\(\Leftrightarrow B=\frac{4x^2+4x-3-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\frac{2x+1}{x^2+2}\)
\(\Leftrightarrow B=\frac{\left(8x-4\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)\left(x^2+2\right)}=\frac{4\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)\left(x^2+2\right)}=\frac{4}{x^2+2}\)
b) \(B=\frac{4}{x^2+2}\left(x\ne\pm\frac{1}{2}\right)\)
Với x=-1 (TMĐK) thay vào B ta có:
\(B=\frac{4}{\left(-1\right)^2+2}=\frac{4}{1+2}=\frac{4}{3}\)
Vậy \(B=\frac{4}{3}\)khi x=-1