Toán Lớp 12: Giải phương trình `5(5x^2-17)^2-343x-833=0` @Khang, đề đúng không sai

Question

Toán Lớp 12:  Giải phương trình 5(5x^2-17)^2-343x-833=0 @Khang, đề đúng không sai

thanhmaianh · Answer

Giải đáp:

5 ( 5x ^2 − 17 )^2 − 343 x − 833 = 0

<=> 125 x^4 – 850 x^2 – 343 x + 612 = 0

<=> 125x^4+175x^3−175x^3−180x^2−245x^2−425x^2+252x−595x+612=0

<=> (125x^4+175x^3−180x^2)−(175x^3+245x^2-252x)−(425x^2+595x-612)=0

<=> 5x^2(25x^2+35x−36)−7x(25x^2+35x-36)−17(25x^2+35x-36)=0

<=> (5 x^2 – 7 x – 17) (25 x^2 + 35 x – 36) = 0

<=> \(\left[ \begin{array}{l}
5 x^2 – 7 x – 17=0
\\
25 x^2 + 35 x – 36 = 0
\end{array} \right.\)

<=> \(\left[ \begin{array}{l}
x^2 – \frac{7 x}{5} – \frac{17}{5} = 0\\
x^2 + \frac{7 x}{5} – \frac{36}{25} = 0\end{array} \right.\)

<=> \(\left[ \begin{array}{l}
x^2 – \frac{7 x}{5} = \frac{17}{5}
\\
x^2 + \frac{7 x}{5} = \frac{36}{25}
\end{array} \right.\)

<=> \(\left[ \begin{array}{l}
x^2 – \frac{7 x}{5} + \frac{49}{100}= \frac{389}{100}
\\
x^2 + \frac{7 x}{5}+\frac{49}{100} = \frac{193}{100}
\end{array} \right.\)

<=> \(\left[ \begin{array}{l}
(x – \frac{7}{10})^2=\frac{389}{100}\\
(x + \frac{7}{10})^2 = \frac{193}{100}
\end{array} \right.\)

<=> \(\left[ \begin{array}{l}
x – \frac{7}{10}=\frac{\sqrt{389}}{10}
\\
x – \frac{7}{10}=\frac{-\sqrt{389}}{10}
\\
x + \frac{7}{10} = \frac{\sqrt{193}}{10}
\\
x + \frac{7}{10} = \frac{-\sqrt{193}}{10}
\end{array} \right.\)

<=> \(\left[ \begin{array}{l}
x =\frac{7+\sqrt{389}}{10}
\\
x =\frac{7-\sqrt{389}}{10}
\\
x = \frac{\sqrt{193}-7}{10}
\\
x = \frac{-\sqrt{193}-7}{10}
\end{array} \right.\)

Vậy phương trình có tập nghiệm:

S= {(7 + sqrt(389))/10 ; (7 – sqrt(389))/10 ; ( sqrt(193)-7)/10 ; ( -sqrt(193)-7)/10}