$$4x^{2}-8x+\sqrt{2x+3}=1$$
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Đã gửi 27-08-2013 - 14:59
Zaraki
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PQT
$$4x^{2}-8x+\sqrt{2x+3}=1$$
Lời giải. Đặt $\sqrt{2x+3}=a$ thì $x= \frac{a^2-3}{2}$. Khi đó phương trình trở thành $$a^4-10a^2+a+20=0 \\ (a^2+a-5)(a^2-a-4)=0$$
Đến đây bạn tự giải tiếp nhé.
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