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\(=x\left(\frac{x^2}{4}+x+1\right)=x\left(\frac{x}{2}+1\right)^2\)
\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
hk
tốt
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
ĐKXĐ: \(x\notin\left\{-7;3;-3\right\}\)
a) Ta có: \(B=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{x-3}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)
\(=\left(\dfrac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10}{x+3}-\dfrac{x+3}{x+3}\right)\)
\(=\dfrac{x^2+1-x^2+3x+5x+15}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+10-x-3}{x+3}\)
\(=\dfrac{8x+16}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+7}\)
\(=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\)
b) Ta có: |x-1|=2
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Thay x=-1 vào biểu thức \(B=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\), ta được:
\(B=\dfrac{8\cdot\left(-1\right)+16}{\left(-1-3\right)\left(-1+7\right)}=\dfrac{-8+16}{-4\cdot6}=\dfrac{8}{-24}=\dfrac{-1}{3}\)
Vậy: Khi x=-1 thì \(B=\dfrac{-1}{3}\)
c) Để \(B=\dfrac{x+5}{6}\) thì \(=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}=\dfrac{x+5}{6}\)
\(\Leftrightarrow6\left(8x+16\right)=\left(x+5\right)\left(x-3\right)\left(x+7\right)\)
\(\Leftrightarrow48x+96=\left(x^2-3x+5x-15\right)\left(x+7\right)\)
\(\Leftrightarrow\left(x^2+2x-15\right)\left(x+7\right)=48x+96\)
\(\Leftrightarrow x^3+7x^2+2x^2+14x-15x-105-48x-96=0\)
\(\Leftrightarrow x^3+9x^2-49x-201=0\)
\(\Leftrightarrow x^3+3x^2+6x^2+18x-67x-201=0\)
\(\Leftrightarrow x^2\left(x+3\right)+6x\left(x+3\right)-67\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+6x-67\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+6x+9-76\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-76\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+3-2\sqrt{19}\right)\left(x+3+2\sqrt{19}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+3-2\sqrt{19}=0\\x+3+2\sqrt{19}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=2\sqrt{19}-3\left(nhận\right)\\x=-2\sqrt{19}-3\left(nhận\right)\end{matrix}\right.\)
Vậy: Để \(B=\dfrac{x+5}{6}\) thì \(x\in\left\{2\sqrt{19}-3;-2\sqrt{19}-3\right\}\)
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