Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH

Lời giải:

Ta có:

\(x^4-4x^3-2x^2+12x+5=0\)

\(\Leftrightarrow (x^4-4x^3+4x^2)-6x^2+12x+5=0\)

\(\Leftrightarrow (x^2-2x)^2-6(x^2-2x)+5=0\)

Đặt $x^2-2x=a$. Khi đó:
\(a^2-6a+5=0\)

\(\Leftrightarrow (a-1)(a-5)=0\Rightarrow \left[\begin{matrix} a=1\\ a=5\end{matrix}\right.\)

Nếu $a=1$ thì $x^2-2x=1$

\(\Leftrightarrow x^2-2x-1=0\Rightarrow x=1\pm \sqrt{2}\)

Nếu $a=5$ thì $x^2-2x=5$

\(\Leftrightarrow x^2-2x-5=0\Rightarrow x=1\pm \sqrt{6}\)

ta có : \(\left\{{}\begin{matrix}\dfrac{8}{3}a+\dfrac{b}{3}=0,1\\\dfrac{8}{3}3b+\dfrac{b}{3}=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{3}a+\dfrac{b}{3}=0,1\\\dfrac{25}{3}b=0,1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{250}\\\dfrac{8}{3}a+\dfrac{\dfrac{3}{250}}{3}=0,1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{250}\\a=\dfrac{9}{250}\end{matrix}\right.\) vậy \(\left(a\overset{.}{,}b\right)=\left(\dfrac{9}{250}\overset{.}{,}\dfrac{3}{250}\right)\)

để phương trình \(5x^2-x+m\le0\) vô nghiệm thì \(5x^2-x+m>0\forall x\) \(\Leftrightarrow\left\{{}\begin{matrix}\Delta< 0\\a>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(-1\right)^2-5\left(m\right)>0\\5>0\left(luônđúng\right)\end{matrix}\right.\) \(\Leftrightarrow1-5m>0\Leftrightarrow m< \dfrac{1}{5}\)

vậy \(m< \dfrac{1}{5}\) thì phương trình \(5x^2-x+m\le0\) vô nghiệm

đkxđ: x≠\(\pm\)2

pt <=> \(\left(x+1\right)\left(x^2-4\right)=x-2\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x+1\right)\left(x+2\right)-1\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+1\right)=0\)

=> \(\left[{}\begin{matrix}x-2=0\\x^2+3x+1=0\end{matrix}\right.\)

+) x-2=0 => x = 2(ktm)

+) \(x^2+3x+1=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\sqrt{\dfrac{5}{4}}\\x+\dfrac{3}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-3}{2}\\x=\dfrac{-\sqrt{5}-3}{2}\end{matrix}\right.\)(t/m)

vậy...........

ta có : \(\Delta'=\left(m-1\right)^2-\left(3m-3\right)=m^2-2m+1-3m+3\)

\(\Leftrightarrow\Delta'=m^2-5m+4\)

để phương trình có 2 nghiệm\(\Leftrightarrow\Delta'\ge0\Leftrightarrow m^2-5m+4\Leftrightarrow\left[{}\begin{matrix}m\ge4\\m\le1\end{matrix}\right.\)

áp dụng định lí vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-1\right)\\x_1x_2=3m-3\end{matrix}\right.\)

ta có : \(x_1^2+x_2^2\ge10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2\ge10\)

\(\Leftrightarrow\left(2\left(m-1\right)\right)^2-2\left(3m-3\right)\ge10\)

\(\Leftrightarrow4m^2-8m+4-6m+6\ge10\)

\(\Leftrightarrow4m^2-14m\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{7}{2}\\m\le0\end{matrix}\right.\) kết hợp với \(\left[{}\begin{matrix}m\ge4\\m\le1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m\ge4\\m\le0\end{matrix}\right.\) vậy \(m\ge4\) hoặc \(m\le0\)