Đáp án:
Giải thích các bước giải:
q)
→$\frac{1}{x}$ +2=x+2x²+ $\frac{1}{x}$ +2
→$\frac{1}{x}$ - $\frac{1}{x}$ -x-2x²=2-2
→-2x²-x=0
→-2x(x+1)=0
→\(\left[ \begin{array}{l}-2x=0\\x+1=0\end{array} \right.\)
→\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
s)
→(x+2)(x-3)(17x²-17x+8)-(x+2)(x-3)(x²-17x+33)=0
→(x+2)(x-3)(17x²-17x+8-x²+17x-33)=0
→(x+2)(x-3)(16x²-25)=0
→(x+2)(x-3)(4x-5)(4x+5)=0
→x=-2;3;$\frac{5}{4}$ ;$\frac{-5}{4}$
r)
→(2x+3)($\frac{3x+8}{2-7x}$ +1)-(x-5)($\frac{3x+8}{2-7x}$ +1)=0
→($\frac{3x+8}{2-7x}$ +1)(2x+3-x+5)=0
→($\frac{3x+8}{2-7x}$ +1)(x+8)=0
→\(\left[ \begin{array}{l}$\frac{3x+8}{2-7x}$ +1=0\\x+8=0\end{array} \right.\)
→\(\left[ \begin{array}{l}$\frac{3x+8}{2-7x}$ +$\frac{2-7x}{2-7x}$ =0\\x=-8\end{array} \right.\)
→\(\left[ \begin{array}{l}3x+8+2-7x=0 \\x=-8\end{array} \right.\)
→\(\left[ \begin{array}{l}-4x+10=0 \\x=-8\end{array} \right.\)
→\(\left[ \begin{array}{l}x=$\frac{5}{2}$ \\x=-8\end{array} \right.\)
