Question

giải bất phương trình sau : 6\(\sqrt{\left(x-2\right)\left(x-32\right)}\)  <= x2 - 34x + 48

 Hướng dẫn : Đặt y = \(\sqrt{\left(x-2\right)\left(x-32\right)}\)  = \(\sqrt{x^2-34x+64}\)

Answer 1

\(6\sqrt{x^2-34x+64}=x^2-34x+48\)

\(\text{đ}at:x^2-34x+48=a\Rightarrow6\sqrt{a+16}=a\Leftrightarrow36a+576=a^2\Leftrightarrow a^2-36a-576=0;\Delta=\left(-36\right)^2-4.\left(-576\right).1=3600\Rightarrow\left\{{}\begin{matrix}a_1=24\\a_2=-96\end{matrix}\right.\)

\(+,a=-96\Rightarrow x^2-34x+48=-96\Leftrightarrow x^2-34x+144=0;\Delta=34^2-4.144=580\Rightarrow\left\{{}\begin{matrix}x_1=-34+2\sqrt{145}\\x_2=-34-2\sqrt{145}\end{matrix}\right.\)

\(+,a=24\Rightarrow x^2-34x+48=24\Leftrightarrow x^2-34x+24=0;\Delta=1156-96=1060\Rightarrow\left\{{}\begin{matrix}x_1=-34+2\sqrt{265}\\x_2=-34-2\sqrt{265}\end{matrix}\right.\)