Question

1-РЕШИТЬ УРАВНЕНИЕ, А 2-УПРОСТИТЬ ВЫРАЖЕНИЕ.

Answer 1

$\dfrac{1}{2x^{2}-3x-9 }+\dfrac{3}{x^{2}-x-6 }=\dfrac{x}{2x^{2} +7x+6} \\\\\\\dfrac{1}{2(x-3)(x+1,5) }+\dfrac{3}{(x-3)(x+2)}-\dfrac{x}{2(x+2)(x+1,5)}=0\\\\\\\dfrac{1\cdot(x+2)+3\cdot(2x+3)-x\cdot(x-3)}{2(x-3)(x+1,5)(x+2)}=0\\\\\\\dfrac{x+2+6x+9-x^{2}+3x }{(x-3)(x+1,5)(x+2)}=0\\\\\\\dfrac{x^{2}-10x-11 }{(x-3)(x+1,5)(x+2) } =0\\\\\\\dfrac{(x-11)(x+1) }{(x-3)(x+1,5)(x+2) } =0\\\\\\\displaystyle\left \{ {{(x-11)(x+1)=0} \atop {x\neq 3 \ ; \ x\neq -1,5 \ ; \ x\neq -2}} \right.$

$x-11=0 \ \Rightarrow \ x_{1}=11\\\\x+1=0 \ \Rightarrow \ x_{2}=-1 \\\\Otvet:\boxed{11 \ ; \ -1}$

$2) \ \dfrac{36a^{2} }{5a^{2} +13a-6}-\dfrac{5a-2}{a+3}= \dfrac{36a^{2} }{5(a+3)(a-0,4)}-\dfrac{5a-2}{a+3} =\\\\=\dfrac{36a^{2}-(5a-2)\cdot5(a-0,4) }{5(a+3)(a-0,4)}=\dfrac{36a^{2}-25a^{2}+20a-4 }{5(a+3)(a-0,4)} =\\\\=\dfrac{11a^{2}+20a-4 }{5(a+3)(a-0,4)} =\dfrac{11(a+2)(a-\frac{2}{11}) }{5(a+3)(a-0,4)} \\\\\\\dfrac{11(a+2)(a-\frac{2}{11}) }{5(a+3)(a-0,4)} :\dfrac{11a-2}{a^{2}-2a-15 }=\dfrac{(a+2)(11a-2) }{5(a+3)(a-0,4)}\cdot\dfrac{(a-5)(a+3)}{11a-2} =\\\\=\dfrac{(a+2)(a-5)}{5a-2}$

$\dfrac{(a+2)(a-5)}{5a-2}-\dfrac{28a-a^{2} }{2-5a}=\dfrac{a^{2}-3a-10 }{5a-2}+\dfrac{28a-a^{2} }{5a-2}= \\\\=\dfrac{a^{2}-3a-10+28a-a^{2}}{5a-2} =\dfrac{25a-10}{5a-2}=\dfrac{5(5a-2)}{5a-2}=\boxed5$