Giải PT trùng phương:
`4,5x^4 + 2,5x ² -2 = 0`(1)
Đặt `x^2=t (t≥0)`
khi đó (1) trở thành `4,5t^2 + 2,5t-2 = 0 <=>`\(\left[ \begin{array}{l}t_1=4/9(N)\\t_2=-1(L)\end{array} \right.\)
Với `t=4/9⇔x^2=4/9⇔`\(\left[ \begin{array}{l}x_1=2/3\\x_2=-2/3\end{array} \right.\)
Vậy `S={2/3;-2/3}`
Giải PT chứa ẩn ở mẫu:
1)`\frac{1}{3x^2-27}+ \frac{3}{4}-\frac{1}{x-3}=1` ĐKXĐ: `xne3`;`xne-3`
`⇔\frac{4}{12(x-3)(x+3)}+ \frac{9(x-3)(x+3)}{12(x-3)(x+3)}-\frac{12(x+3)}{12(x-3)(x+3)}=\frac{12(x-3)(x+3)}{12(x-3)(x+3)}`
`⇒4+9(x-3)(x+3)-12(x+3)=12(x-3)(x+3)`
`⇔4+9(x^2-9)-12(x+3)=12(x^2-9)`
`⇔4+9x^2-81-12x-26=12x^2-108`
`⇔ 9x^2 -12x -12x^2=-108-4+26+81`
`⇔ -3x^2 -12x +5 =0`
`⇔ 3x^2 +12x -5 =0`
Ta có: `Δ'=36+15=51`
\(\left[ \begin{array}{l}x_1=\dfrac{-6+\sqrt {51}}{3}\\x_2=\dfrac{-6-\sqrt {51}}{3}\end{array} \right.\)
Vậy `S={\frac{-6+\sqrt {51}}{3};\frac{-6-\sqrt {51}}{3}}`
2) `\frac{30}{x^2-1}-\frac{13}{x^2+x+1}=\frac{18x+7}{x^3-1}` ĐKXĐ: `xne±1`
`⇔\frac{30}{(x-1)(x+1)}-\frac{13}{x^2+x+1}=\frac{18x+7}{(x-1)(x^2+x+1)}`
`⇔\frac{30(x^2+x+1)}{(x-1)(x+1)(x^2+x+1)}-\frac{13(x+1)(x-1)}{(x+1)(x-1)(x^2+x+1)}=\frac{(18x+7)(x+1)}{(x+1)(x-1)(x^2+x+1)}`
`⇒30x^2+30x+30-13x^2+13=18x^2+18x+7x+7`
`⇔30x^2+30x -13x^2 -18x^2-18x-7x=-13+7-30`
`⇔-x^2+5x+36=0`
Ta có: `Δ=25+144=169`
`<=>`\(\left[ \begin{array}{l}x_1=-4(N)\\x_2=9(N) \end{array} \right.\)
Vậy `S={-4;9}`
