Question

розв'яжіть рівняння ​

Answer 1

$\dfrac{10\,x}{\left(x+1\right)\,\left(x+4\right)}+\dfrac{9\,x}{\left(x+2\right)^{2}}=2\\0\,x\,\left(x^{2}+4\,x+4\right)+9\,x\,\left(x^{2}+5\,x+4\right)-2\,\left(x^{4}+9\,x^{3}+28\,x^{2}+36\,x+16\right)=0\\-2\,x^{4}+x^{3}+29\,x^{2}+4\,x-32=0\Leftrightarrow 2\,x^{4}-x^{3}-29\,x^{2}-4\,x+32=0\\\left(2\,x^{2}+\dfrac{32}{x^{2}}\right)+\left(-x-\dfrac{4}{x}\right)-29=0\Leftrightarrow 2\,\left(x^{2}+\dfrac{16}{x^{2}}\right)-1\,\left(x+\dfrac{4}{x}\right)-29=0$$2\,\left(x^{2}+8+\dfrac{16}{x^{2}}-8\right)-1\,\left(x+\dfrac{4}{x}\right)-29=0\Leftrightarrow 2\cdot \left(x+\dfrac{4}{x}\right)^{2}-1\,\left(x+\dfrac{4}{x}\right)-45=0\\x+\dfrac{4}{x}=t\Rightarrow 2\,t^{2}-t-45=0\Rightarrow t=\left \{ -\frac{9}{2},5 \right \}\\x+\dfrac{4}{x}=5\Leftrightarrow x^{2}-5\,x+4=0\Rightarrow x=\left \{ 1,4 \right \}\\x+\dfrac{4}{x}=-\dfrac{9}{2}\Leftrightarrow x^{2}+\dfrac{9}{2}x+4=0\Rightarrow x=\left \{\pm \dfrac{\sqrt{17}}{4}-\dfrac{9}{4} \right \}$