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a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
c/
\(x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x=t\)
\(t\left(t+2\right)-24=0\Leftrightarrow t^2+2t-24=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+3x=4\\x^2+3x=-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4-2x^3+x^2+3x^2-3x-10=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+3\left(x^2-x\right)-10=0\)
Đặt \(x^2-x=t\)
\(t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x=2\\x^2-x=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\end{matrix}\right.\)
a/ ĐKXĐ: ...
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(2\left(t^2-2\right)-3t+2=0\)
\(\Leftrightarrow2t^2-3t-2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=2\\x+\frac{1}{x}=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-2x=1=0\\2x^2-x+2=0\end{matrix}\right.\)
b/ Với \(x=0\) ko phải nghiệm
Với \(x\ne0\) chia 2 vế của pt cho \(x^2\)
\(x^2+\frac{1}{x^2}-5x+\frac{5}{x}-8=0\)
\(\Leftrightarrow x^2+\frac{1}{x^2}-2-5\left(x-\frac{1}{x}\right)-6=0\)
Đặt \(x-\frac{1}{x}=t\Rightarrow t^2=x^2+\frac{1}{x^2}-2\)
\(t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{x}=-1\\x-\frac{1}{x}=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x-1=0\\x^2-6x-1=0\end{matrix}\right.\)
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)
\(1>=\left(x+y\right)^2>=\left(2\sqrt{xy}\right)^2=4xy\Rightarrow1>=4xy\Rightarrow\frac{1}{2}>=2xy\)(bđt cosi)
\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}>=\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{1}{2}}\)
\(=\frac{4}{\left(x+y\right)^2}+2>=\frac{4}{1^2}+2=4+2=6\)
dấu = xảy ra khi \(x=y=\frac{1}{2}\)
vậy min \(\frac{1}{x^2+y^2}+\frac{1}{xy}=6\)khi \(x=y=\frac{1}{2}\)
E = { x ∈ N / 3x - 24 < 0 }
F = { x ∈ N* / 2 - 5x > \(\frac{X}{2}\) }
G = { x ∈ N / x4 - 4x2 + 3 < 0 }