Question

решите пожалуйста уравнение с использованием переменной t

Answer 1

x = 10 или x = 8/11 или х = - 6/13 или х = 0

всё есть на фото

хорошего дня

Answer 2

$\Big(\dfrac{5x+1}{2x-3}\Big)^{2} +\Big(\dfrac{3-2x}{5x+1}\Big)^{2} =\dfrac{82}{9}\\\\\dfrac{5x+1}{2x-3} =t \ \Rightarrow \ \dfrac{3-2x}{5x+1} =\dfrac{82}{9}\\\\t^{2} +\dfrac{1}{t^{2} }=\dfrac{82}{9}\\\\\Big(t+\dfrac{1}{t}\Big)^{2}-2=\dfrac{82}{9}\\\\\Big(t+\dfrac{1}{t}\Big)^{2}=\dfrac{100}{9}$

$\left[\begin{array}{ccc}t+\dfrac{1}{t}=-\dfrac{10}{3} \\\\t+\dfrac{1}{t}=\dfrac{10}{3} \end{array}\right \\\\1) \ t+\dfrac{1}{t} =-\dfrac{10}{3}\\\\t+\dfrac{1}{t} +\dfrac{10}{3}=0, \ t\neq 0\\\\3t^{2}+10t+3=0\\\\D=10^{2}-4\cdot 3\cdot 3=100-36=64=8^{2}\\\\t_{1}=\dfrac{-10-8}{6} =-3\\\\t_{2}=\dfrac{-10+8}{6} =-\dfrac{1}{3} \\\\\dfrac{5x+1}{2x-3}=-3 \\\\\dfrac{5x+1}{2x-3}+3=0, \ x\neq 1,5\\\\5x+1+6x-9=0\\\\11x=8\\\\x_{1}=\dfrac{8}{11}$

$\dfrac{5x+1}{2x-3}=-\dfrac{1}{3}\\\\\dfrac{5x+1}{2x-3}+\dfrac{1}{3} =0\\\\15x+3+2x-3=0\\\\17x=0\\\\x_{2}=0$

$2) \ t+\dfrac{1}{t} =\dfrac{10}{3}\\\\t+\dfrac{1}{t} -\dfrac{10}{3}=0, \ t\neq 0\\\\3t^{2}-10t+3=0\\\\D=(-10)^{2}-4\cdot 3\cdot 3=100-36=64=8^{2}\\\\t_{1}=\dfrac{10-8}{6} =\dfrac{1}{3} \\\\t_{2}=\dfrac{10+8}{6} =3 \\\\\dfrac{5x+1}{2x-3}=3 \\\\\dfrac{5x+1}{2x-3}-3=0, \ x\neq 1,5\\\\5x+1-6x+9=0\\\\-x=-10\\\\x_{3}=10$

$\dfrac{5x+1}{2x-3}=\dfrac{1}{3} \\\\\dfrac{5x+1}{2x-3}-\dfrac{1}{3} =0\\\\15x+3-2x+3=0\\\\13x=-6\\\\x_{4} =-\dfrac{6}{13}$