`~rai~`

\(e)(5x-4)^2-49x^2\\=(5x-4)^2-(7x)^2\\=(5x-4+7x)(5x-4-7x)\\=(12x-4)(-2x-4)\\=4.(3x-1).(-2).(x+2)\\=-8(3x-1)(x+2).\\\text{Xét }\\+)-8(3x-1)(x+2)>0\\\Leftrightarrow (3x-1)(x+2)<0\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}3x-1>0\\x+2<0\end{cases}\\\begin{cases}3x-1<0\\x+2>0\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}x>\dfrac{1}{3}\\x<-2\end{cases}\text{(vô lí)}\\\begin{cases}x<\dfrac{1}{3}\\x>-2\end{cases}\end{array}\right.\\\Leftrightarrow -2<x<\dfrac{1}{3}.\\+)-8(3x-1)(x+2)<0\\\Leftrightarrow (3x-1)(x+2)>0\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}3x-1>0\\x+2>0\end{cases}\\\begin{cases}3x-1<0\\x+2<0\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}x>\dfrac{1}{3}\\x>-2\end{cases}\\\begin{cases}x<\dfrac{1}{3}\\x<-2\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x>\dfrac{1}{3}\\x<-2.\end{array}\right.\\+)-8(3x-1)(x+2)=0\\\Leftrightarrow (3x-1)(x+2)=0\\\Leftrightarrow \left[\begin{array}{I}3x-1=0\\x+2=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{1}{3}\\x=-2.\end{array}\right.\\\text{Vậy giá trị biểu thức:}\\\text{+)Dương khi }-2<x<\dfrac{1}{3};\\\text{+)Âm khi }x>\dfrac{1}{3}\text{ hoặc }x<-2;\\\text{+)Bằng 0 khi x}\in\left\{-2;\dfrac{1}{3}\right\}.\\\\f)(2x+5)^2-(x-9)^2\\=[(2x+5)+(x-9)][(2x+5)-(x-9)]\\=(2x+5+x-9)(2x+5-x+9)\\=(3x-4)(x+14).\\\text{Xét:}\\+)(3x-4)(x+14)>0\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}3x-4>0\\x+14>0\end{cases}\\\begin{cases}3x-4<0\\x+14<0\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}x>\dfrac{4}{3}\\x>-14\end{cases}\\\begin{cases}x<\dfrac{4}{3}\\x<-14\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x>\dfrac{4}{3}\\x<-14.\end{array}\right.\\+)(3x-4)(x+14)<0\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}3x-4>0\\x+14<0\end{cases}\\\begin{cases}3x-4<0\\x+14>0\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}x>\dfrac{4}{3}\\x<-14\end{cases}\text{(vô lí)}\\\begin{cases}x<\dfrac{4}{3}\\x>-14\end{cases}\end{array}\right.\\\Leftrightarrow -14<x<\dfrac{4}{3}.\\+)(3x-4)(x+14)=0\\\Leftrightarrow \left[\begin{array}{I}3x-4=0\\x+14=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{4}{3}\\x=-14.\end{array}\right.\\\text{Vậy giá trị biểu thức:}\\\text{+)Dương khi }x<\dfrac{4}{3}\text{ hoặc }x<-14.\\\text{+)Âm khi }-14<x<\dfrac{4}{3}.\\\text{+)Bằng 0 khi x}\in\left\{-14;\dfrac{4}{3}\right\}.\)