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Lời giải:
a) Nếu không điều kiện gì của $x$ thì biểu thức không có GTNN
vì cho $x$ chạy từ \(-100\) đến âm vô cùng thì giá trị $A$ càng nhỏ (âm) vô cùng
b) Điều kiện: \(x>0\)
\(B=\frac{\left ( x+\frac{1}{x} \right )^6-\left ( x^6+\frac{1}{x^6} \right )-2}{\left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right )}=\frac{\left ( x+\frac{1}{x} \right )^6-\left [ (x^3+\frac{1}{x^3})^2-2 \right ]-2}{\left ( x+\frac{1}{x}\right )^3+\left ( x^3+\frac{1}{x^3} \right )}\)
\(=\frac{\left ( x+\frac{1}{x} \right )^6-\left ( x^3+\frac{1}{x^3} \right )^2}{\left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right )}=\frac{\left [ \left ( x+\frac{1}{x} \right )^3-\left ( x^3+\frac{1}{x^3} \right ) \right ]\left [ \left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right ) \right ]}{\left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right )}\)
\(=\left ( x+\frac{1}{x} \right )^3-\left ( x^3+\frac{1}{x^3} \right )=\left ( x+\frac{1}{x} \right )^3-\left [ \left ( x+\frac{1}{x} \right )^3-3.x.\frac{1}{x}\left ( x+\frac{1}{x} \right ) \right ]\) (sd hằng đẳng thức đáng nhớ \(x^3+y^3=(x+y)^3-3xy(x+y)\) )
\(=3\left(x+\frac{1}{x}\right)\geq 3.2\sqrt{x.\frac{1}{x}}=6\) (theo BĐT Cô-si cho hai số dương)
Vậy \(B_{\min}=6\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x=\frac{1}{x}\\ x>0\end{matrix}\right.\Leftrightarrow x=1\)
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
a/ \(\dfrac{x-1}{x+1}-\dfrac{x}{x+2}=\dfrac{x-3}{\left(x+1\right)\left(x+2\right)}\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)-x\left(x+1\right)=x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x=x-3\)
\(\Leftrightarrow-x=-1\Leftrightarrow x=1\left(tm\right)\)
Vậy...............................
b/ \(\dfrac{x-3}{2}\ge0\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)
Vậy....................
c/ ĐK: x khác 2
\(\dfrac{x+1}{x-2}>0\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\left(tm\right)\)
Vậy.................
d/ \(x^2-x-6\ge0\)
\(\Leftrightarrow x^2-3x+2x-6\ge0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\le-2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-2\end{matrix}\right.\)
Vậy x ≥ 3 hoặc x ≤-2