\(3,\\ a,=3\left(x^2+2xy-z^2+y^2\right)=3\left[\left(x+y\right)^2-z^2\right]\\ =3\left(x+y-z\right)\left(x+y+z\right)\\ b,=\left[5\left(x-5\right)-\left(x+3\right)\right]\left[5\left(x-5\right)+\left(x+3\right)\right]\\ =\left(5x-25-x-3\right)\left(5x-25+x+3\right)\\ =\left(4x-28\right)\left(6x-22\right)=4\left(x-7\right)2\left(3x-11\right)\\ =8\left(x-7\right)\left(3x-11\right)\\ 4,\\ a,\Rightarrow x\left(49x^2-36\right)=0\\ \Rightarrow x\left(7x-6\right)\left(7x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{6}{7}\\x=-\dfrac{6}{7}\end{matrix}\right.\\ b,\Rightarrow x\left(3x-1\right)+\dfrac{3}{4}\left(1-3x\right)=0\\ \Rightarrow\left(x-\dfrac{3}{4}\right)\left(3x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)
