ПОМОГИТЕ,СРОЧНО ЛЮДИ(((

$4x^4-17x^2-15=0 \\ t = x^2 \\ 4t^2-17t-15=0 \\ D = 289-4*4*-15 = 529 \\ x_1,x_2 = \frac{17+-23}{8} \\ t_1 = 5 ; t_2 = -6/8 \\ x^2 = 5 ; t_2 = wrong-answer \ \textless \ 0. \\ x = +-\sqrt{5} \\$
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$3x-(2x-7) \leq 3(1+x) \\ 3x-2x+7 \leq 3+3x \\ 7-3 \leq 5x-3x \\ 4 \leq 2x |:2 \\ 2 \leq x$
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$\left \{ {{3x+17\ \textless \ 2} \atop {3-4x\ \textless \ 19}} \right. \\ 3x\ \textless \ -15 |:3 \\ x\ \textless \ -5 \\ 3-4x\ \textless \ 19 \\ 3-19\ \textless \ 4x \\ -16\ \textless \ 4x |:4 \\ -4\ \textless \ x \\ Answer: there-is-no-solution$ $\frac{x+3}{7} = \frac{2-x}{3} \\ 3(x+3) = 7(2-x) \\ 3x+9=14-7x \\ 10x = 5 |:10 \\ x = 0.5$
$(x-1/3)(x-1/5) \leq 0 \\ x = 1/3 ; x = 1/5 \\ 1/5 \leq x \leq 1/3$