Gọi \(M\left(x;y;z\right)\Rightarrow\overrightarrow{BM}=\left(x;y;z-3\right)\)

\(cos\left(\widehat{\overrightarrow{BM};\overrightarrow{Oz}}\right)=cos30^o=\dfrac{\left|z-3\right|}{\sqrt{x^2+y^2+\left(z-3\right)^2}}=\dfrac{\sqrt{3}}{2}\)

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

\(\Leftrightarrow3\left[x^2+y^2+\left(z-3\right)^2\right]=4\left(z-3\right)^2\)

\(\Leftrightarrow3x^2+3y^2=\left(z-3\right)^2\)

\(\Leftrightarrow z-3=\pm\sqrt{3x^2+3y^2}\left(1\right)\)

\(AM=\sqrt{\left(x-1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2}\)

\(\left(1\right)\Leftrightarrow AM=\sqrt{\left(x-1\right)^2+\left(y-2\right)^2+\left(\sqrt{3x^2+3y^2}\right)^2}=\sqrt{\left(x-1\right)^2+\left(y-2\right)^2+3x^2+3y^2}\)

\(\Leftrightarrow AM=\sqrt{x^2-2x+1+y^2-4y+4+3x^2+3y^2}\)

\(\Leftrightarrow AM=\sqrt{4x^2-2x+4y^2-4y+5}\left(2\right)\)

Đặt \(f\left(x;y\right)=4x^2-2x+4y^2-4y+5\)

\(f'_x=8x-2=0\Leftrightarrow x=\dfrac{1}{4}\)

\(f'_y=8y-4=0\Leftrightarrow y=\dfrac{1}{2}\)

\(\Rightarrow\) Điểm dừng \(M\left(\dfrac{1}{4};\dfrac{1}{2};z\right)\)

\(\left(1\right)\Rightarrow z=\pm\dfrac{\sqrt{15}}{3}+3\Rightarrow M\left(\dfrac{1}{4};\dfrac{1}{2};\pm\dfrac{\sqrt{15}}{4}+3\right)\)

\(f''_{xx}=8;f''_{yy}=8;f''_{xy}=0\)

Ma trận Hessian:\(H=\left[\begin{matrix}8&0\\0&8\end{matrix}\right]\) \(\Rightarrow\left\{{}\begin{matrix}H_1=8>0\\H_2=64-0=64>0\end{matrix}\right.\)

\(\Rightarrow M\left(\dfrac{1}{4};\dfrac{1}{2};\pm\dfrac{\sqrt{15}}{4}+3\right)\) là điểm cực tiểu của \(f\left(x;y\right)\)

\(\left(2\right)\Rightarrow AM\left(min\right)=\sqrt{4\left(\dfrac{1}{4}\right)^2-2\left(\dfrac{1}{4}\right)+4\left(\dfrac{1}{2}\right)^2+5}=\dfrac{\sqrt{15}}{2}\approx1,94\left(km\right)\)

Vậy \(AM=1,94\left(km\right)\) thỏa mãn đề bài

Bài 16 :

Xét \(S\left(x\right)=A\left(x\right)+B\left(x\right)+C\left(x\right)\)

\(\Rightarrow S\left(x\right)=16x^4-8x^3+7x^2-9-15x^4+3x^3-5x^2-6+5x^3+3x^2+18\)

\(\Rightarrow S\left(x\right)=x^4+5x^2+3\)

mà \(\left\{{}\begin{matrix}x^4\ge0\\5x^2\ge0\\3>0\end{matrix}\right.\) \(\forall x\in R\)

\(\Rightarrow S\left(x\right)=x^4+5x^2+3>0;\forall x\in R\)

\(\Rightarrow\) Có ít nhất 1 trong 3 đa thức \(A\left(x\right);B\left(x\right);C\left(x\right)\) phải dương với mọi \(x\)

\(\Rightarrowđpcm\)

a) \(y=-2x+1\left(d\right)\)

Hệ số góc \(a=-2\)

 \(a=tan\left(\widehat{\left(d\right);Ox}\right)=-2< 0\)

\(\Rightarrow90^o< \left(\widehat{\left(d\right);Ox}\right)< 180^o\) là góc tù

b) \(\left(d'\right):y=-2x-3\Rightarrow HSG.a=-2\)

\(\left(d\right)//\left(d'\right):y=ax+b\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b\ne-3\end{matrix}\right.\)

c) \(y=-2x+1\)

a) \(y=\dfrac{1}{2}x+1\left(d_1\right)\)

b) Để \(\left(d_1\right)\) cắt \(\left(d_2\right)\)  khi và chỉ khi

\(\Leftrightarrow2-3m\ne\dfrac{1}{2}\)

\(\Leftrightarrow3m\ne\dfrac{3}{2}\)

\(\Leftrightarrow m\ne\dfrac{1}{2}\) thỏa mãn đề bài

Ta thấy \(\Delta'=\left(m+1\right)^2-2m=m^2+1>0;\forall m\in R\)

\(\Rightarrow\left(1\right)\) luôn luôn có 2 nghiệm phân biệt \(x_1;x_2\)

Theo định lý Vi-et ta có : \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{matrix}\right.\)

Theo đề bài ta có :

\(x_1^2+x_2^2=\left(2\sqrt{5}\right)^2=20\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=20\)

\(\Leftrightarrow4\left(m+1\right)^2-2.2m=20\)

\(\Leftrightarrow4m^2+4m-16=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{-1+\sqrt{17}}{2}\\m=\dfrac{-1-\sqrt{17}}{2}\end{matrix}\right.\) thỏa mãn đề bài

\(\left(\alpha\right):\left(2m-1\right)x-3my+2z+3=0\Rightarrow\overrightarrow{n_1}=\left(2m-1;-3m;2\right)\)

\(\left(\beta\right):mx+\left(m-1\right)y+4z-5=0\Rightarrow\overrightarrow{n_2}=\left(m;m-1;4\right)\)

\(\left(\alpha\right)\perp\left(\beta\right)\Rightarrow\overrightarrow{n_1}.\overrightarrow{n_2}=0\)

\(\Leftrightarrow\left(2m-1\right)m-3m\left(m-1\right)+2.4=0\)

\(\Leftrightarrow2m^2-m-3m^2+3m+8=0\)

\(\Leftrightarrow m^2-2m-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-2\\m=4\end{matrix}\right.\)

Vậy với \(m=-2\cup m=4\) thỏa mãn yêu cầu đề bài

\(\dfrac{y+z+3}{x}=\dfrac{x+z+4}{y}=\dfrac{x+y-7}{z}=\dfrac{y+z+3+x+z+4+x+y-7}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+3}{x}=\dfrac{\dfrac{1}{2}-x+3}{x}=2\\\dfrac{z+z+4}{y}=\dfrac{\dfrac{1}{2}-y+4}{y}=2\\\dfrac{x+y-7}{z}=\dfrac{\dfrac{1}{2}-z-7}{z}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.\dfrac{1}{3}=\dfrac{5}{6}\\y=\dfrac{9}{2}.\dfrac{1}{3}=\dfrac{3}{2}\\z=\dfrac{\left(-13\right)}{2}.\dfrac{1}{3}=-\dfrac{13}{6}\end{matrix}\right.\)

\(\Rightarrow A=\left(6.\dfrac{7}{6}-6\right)^{2023}+\left(2.\dfrac{3}{2}-2\right)^{2024}+\left(6.\left(\dfrac{-13}{6}\right)+12\right)^{2022}=1^{2023}+1^{2024}+\left(-1\right)^{2022}=3\)

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2010}\right)\)

\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2009}{2010}=\dfrac{1}{2010}\)

Câu 17 :

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)

\(\Rightarrow\left(\dfrac{x}{a}\right)^2=\left(\dfrac{y}{b}\right)^2=\left(\dfrac{z}{c}\right)^2=\left(x+y+z\right)^2\left(1\right)\)

\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\)

a) \(...A\left(x\right)=x^4-3x+3\)

\(...B\left(x\right)=x^4+5x^3+x+3\)

b) \(A\left(2\right)=2^4-3.2+3=16-6+3=13\)

c) \(A\left(x\right)+B\left(x\right)=2x^4+5x^3-2x+6\)

\(A\left(x\right)-B\left(x\right)=-5x^3-4x\)