b)$(2x^2-3x+1)(2x^2+5x+1)=9x^2$+)$x=0$ khong la nghiem+)$x\neq 0: $ chia ca 2 ve cho $x^2$:pt$\Leftrightarrow (x-3+\frac{1}{x})(x+5+\frac{1}{x})=9$Dat $a=x+\frac{1}{x}$ pt tro thanh :$a^2+2a-24=0$ tu giai tiep nha':)
b)$(2x^2-3x+1)(2x^2+5x+1)=9x^2$+)$x=0$ khong la nghiem+)$x\neq 0: $ chia ca 2 ve cho $x^2$:pt$\Leftrightarrow (x-3+\frac{1}{x})(x+5+\frac{1}{x})=9$Dat $a=x+\frac{1}{a}$ pt tro thanh :$a^2+2a-24=0$ tu giai tiep nha':)
b)$(2x^2-3x+1)(2x^2+5x+1)=9x^2$+)$x=0$ khong la nghiem+)$x\neq 0: $ chia ca 2 ve cho $x^2$:pt$\Leftrightarrow (x-3+\frac{1}{x})(x+5+\frac{1}{x})=9$Dat $a=x+\frac{1}{
x}$ pt tro thanh :$a^2+2a-24=0$ tu giai tiep nha':)