§3. Phương trình và hệ phương trình bậc nhất nhiều ẩn

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt[3]{x+6}-2+\sqrt{x-1}-1=x^2-4\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{x-2}{\sqrt[]{x-1}+1}=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}=x+2\left(1\right)\end{matrix}\right.\)

Xét (1), do \(x\ge1\Rightarrow\left\{{}\begin{matrix}x+2\ge3\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}< \dfrac{1}{4}+\dfrac{1}{1}< 3\\\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)

\(\Delta'=\left(m-1\right)^2-3\left(m-2\right)=m^2-5m+7=\left(m-\dfrac{5}{2}\right)^2+\dfrac{3}{4}>0;\forall m\)

\(\Rightarrow\) Phương trình đã cho luôn có 2 nghiệm pb

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=3\left(m-2\right)\end{matrix}\right.\)

Kết hợp Viet và đề bài: \(\left\{{}\begin{matrix}x_1+2x_2=2\\x_1+x_2=2m-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_2=4-2m\\x_1+x_2=2m-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=4m-6\\x_2=4-2m\end{matrix}\right.\)

Thế vào \(x_1x_2=3\left(m-2\right)\)

\(\Rightarrow\left(4m-6\right)\left(4-2m\right)=3\left(m-2\right)\)

\(\Rightarrow\left(m-2\right)\left(8m-9\right)=0\Rightarrow\left[{}\begin{matrix}m=2\\m=\dfrac{9}{8}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+y=m\\2x+my=2m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y=2m\\2x+my=2m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2-m\right)y=-2\\x+y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{m-2}\\x=\dfrac{m^2-2m-2}{m-2}\end{matrix}\right.\)

Vậy ..

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+z=2\\4y-4z=-6\\-y+z=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+z=2\\y-z=-\dfrac{3}{2}\\-y+z=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+z=2\\y-z=-\dfrac{3}{2}\\0=-\dfrac{29}{2}\end{matrix}\right.\)

Hệ đã cho vô nghiệm