x - 2016 / ( -9 ) <0
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\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)
\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)
\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)
hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)
\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)
\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)
\(< =>\left(x^2-9\right)^2-\left[3\left(x-3\right)\right]^2=0\)
\(< =>\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)
\(< =>\left(x^2-9+3x-9\right)\left(x^2-9-3x+9\right)=0\)
\(< =>\left(x^2+3x-18\right)\left(x^2-3x\right)=0\)
\(=>\left[{}\begin{matrix}x^2+3x-18=0\\x^2-3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}\left(x+6\right)\left(x-3\right)=0\\x\left(x-3\right)=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=-6\\x=3\\x=0\end{matrix}\right.\)
\(a,\Leftrightarrow x^2-2x-x^2+5x=6\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=2\)
\(b,\Leftrightarrow x^2-6x+9-x+9=0\\ \Leftrightarrow x^2-7x+18=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{23}{4}=0\\ \Leftrightarrow\left(x-\dfrac{7}{2}\right)^2+\dfrac{23}{4}=0\left(vôlí\right)\)
Ta có: \(2015^{2016}=2015^{2000}.2015^{16}\)
Và \(2016^{2015}=2016^{2000}.2016^{15}\)
=> Ta có: \(2015^{2000}< 2016^{2000}\)
\(2015^{16}< 2016^{15}\)
Vậy \(2015^{2016}< 2016^{2015}\)
x - 2016 / (-9) < 0
Mà -9 < 0
=> x - 2016 > 0
=> x > 2016