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\(4x-2y-4x^2+4xy-y^2\)
\(=\left(4x-2y\right)-\left(4x^2-4xy+y^2\right)\)
\(=2\left(2x-y\right)-\left(2x-y\right)^2\)
\(=\left(2x-y\right)\left(2-2x+y\right)\)
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Lời giải:
Vì $x=9$ nên $x-9=0$
Ta có:
$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$
$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$
$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$
$=x-10=9-10=-1$
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\(9\left(x+3\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x+9\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x+9+2x-1\right)\left(3x+9-2x+1\right)=0\)
\(\Leftrightarrow\left(5x+8\right)\left(x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+8=0\\x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-8}{5}\\x=-10\end{cases}}\)
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\(^{\left(x^2-9\right)^2-9\left(x-3\right)^2}=0\)
\(\Leftrightarrow\)\(^{[\left(x^{ }-3\right)\left(x+3\right)]^2-9\left(x-3\right)^2}=0\)
\(\Leftrightarrow\)\(^{[\left(x^{ }-3\right)]^2[\left(x+3\right)^2-9]}=0\)
\(\Leftrightarrow\)(x-3 )^2=0 hoặc (x+3)^2=9
\(\Leftrightarrow\)x-3 =0 hoặc x+3=3 hoặc x+3=-3
\(\Leftrightarrow\) x=3 hoặc x=0 hoặc x=-6
Vậy x thuộc { -6; 0; 3}