\(\left(a^2+b^2+c^2+1\right)x=ab+bc+ca\)

\(\Leftrightarrow x=\dfrac{ab+bc+ca}{a^2+b^2+c^2+1}\)

Ta có:

\(x^2-1=\dfrac{\left(ab+bc+ca\right)^2}{\left(a^2+b^2+c^2+1\right)^2}-1=\dfrac{\left(ab+bc+ca-a^2-b^2-c^2-1\right)\left(ab+bc+ca+a^2+b^2+c^2+1\right)}{\left(a^2+b^2+c^2+1\right)^2}\)

\(=\dfrac{\left[-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2-2\right]\left[\left(a+b+c\right)^2+a^2+b^2+c^2+2\right]}{4\left(a^2+b^2+c^2+1\right)^2}< 0\)

\(\Rightarrow x^2-1< 0\Rightarrow\left|x\right|< 1\)

Bài 1.

\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=m-1\end{matrix}\right.\)

\(x_0^2+y_0^2=9m\)

\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)

\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)

\(\Leftrightarrow2m^2-7m+5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )

\(\left\{{}\begin{matrix}x-2y=3-m\\2x+y=3m+6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=3-m\\4x+2y=6m+12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=3-m\\5x=5m+15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=m\end{matrix}\right.\)

\(A=\left(m+3\right)^2+m^2=2m^2+6m+9=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(m+\dfrac{3}{2}=0\Rightarrow m=-\dfrac{3}{2}\)

Mặt cầu tâm \(I\left(2;1;1\right)\) bán kính \(R=3\)

Xét mặt phẳng (P) chứa M có phương trình: \(x+2y+2z-A=0\)

Ta cần tìm A nhỏ nhất sao cho (P) cắt (S) tại ít nhất 1 điểm

\(\Rightarrow d\left(I;\left(P\right)\right)\le R\Leftrightarrow\frac{\left|2+2+2-A\right|}{\sqrt{1^2+2^2+2^2}}\le3\)

\(\Leftrightarrow\left|A-6\right|\le9\Rightarrow-9\le A-6\le9\Rightarrow-3\le A\le15\)

\(\Rightarrow A_{min}=-3\Rightarrow\) phương trình (P): \(x+2y+2z+3=0\)

Pt đường thẳng d qua I và vuông góc (P): \(\left\{{}\begin{matrix}x=2+t\\y=1+2t\\z=1+2t\end{matrix}\right.\)

M là giao điểm (P) và d nên tọa độ thỏa mãn:

\(2+t+2\left(1+2t\right)+2\left(1+2t\right)+3=0\Rightarrow t=-1\Rightarrow M\left(1;-1;-1\right)\)

\(\Rightarrow x+y+z=-1\)

• \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {1 + x} \right) = 1 + 1 = 2\).

\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} 1 = 1\).

Vì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ - }} {\rm{ }}f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).

• \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {5 - x} \right) = 5 - 2 = 3\).

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {1 + x} \right) = 1 + 2 = 3\).

Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} {\rm{ }}f\left( x \right) = 3\) nên \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 3\).

Ta có: \(f\left( 2 \right) = 1 + 2 = 3\).

Vậy \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\).