Xét \(\Delta ABO':\)

\(AB\ge O'A-O'B\left(1\right)\)

Xét \(\Delta OAO':\)

\(O'A\ge O'O-OA\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow AB\ge O'O-OA-O'B=950-500-300=150\left(m\right)\)

Dấu '=' xảy ra khi \(4\) điểm \(O;A;B;O'\) thẳng hàng

\(\Rightarrow\) Xây cầu có chiều dài là \(150\left(m\right)\) trên đoạn nối 2 tâm cầu 2 hòn đảo (O'O) thì cây cầu sẽ ngắn nhất.

a) Sửa lại đề bài \(xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+3xyz\)

 \(=xy\left(x+y\right)+xyz+yz\left(y+z\right)+xyz+zx\left(z+x\right)++xyz\)

\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+zx\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(xy+yz+zx\right)\)

b) Đặt \(t=a-2\Rightarrow\left\{{}\begin{matrix}3t-1=3a-7\\3t+1=3a-5\end{matrix}\right.\)

\(...=t\left(3t-1\right)\left(3t+1\right)-8\)

\(=t\left(9t^2-1\right)-8\)

\(=9t^3-t-8\)

\(=9t^3-9t+8t-8\)

\(=9\left(t^3-1\right)+8\left(t-1\right)\)

\(=9\left(t-1\right)\left(t^2+t+1\right)+8\left(t-1\right)\)

\(=\left(t-1\right)\left[9\left(t^2+t+1\right)+8\right]\)

\(=\left(t-1\right)\left(9t^2+9t+17\right)\)

\(=\left(a-3\right)\left[9\left(a-2\right)^2+9\left(a-2\right)+17\right]\)

\(2=1+\sqrt{1}\Rightarrow2=1+\sqrt{-1}.\sqrt{-1}\left(Sai\right)\)

\(\dfrac{A}{B}=\dfrac{-x^{3n}y^2z^{3n-7}}{5x^3y^{2n-6}z^n}=-\dfrac{1}{5}x^{3n-3}y^{-2n+8}z^{2n-7}\)

Để \(A⋮B\) khi và chỉ khi

\(\Leftrightarrow\left\{{}\begin{matrix}3n-3\ge0\\-2n+8\ge0\\2n-7\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n\ge1\\n\le4\\n\ge\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow n=4\left(n\in N\right)\)

Trả lời \(n=4\)

a) \(...\Rightarrow x\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)

b) \(...\Rightarrow x\left(x^3-2x^2+10x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^3-2x^2+10x-20=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+10=0\left(vô.lý\right)\end{matrix}\right.\Leftrightarrow x=2\)

Vậy \(x\in\left\{0;2\right\}\)

c) \(...\Rightarrow\left[{}\begin{matrix}2x-3=x+5\\2x-3=-x-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

d) \(...\Rightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(v_1=15\left(m/s\right)=54\left(km/h\right)\)

\(v_2=36\left(km/h\right)\)

\(t_1=3\left(h\right);t_2=2\left(h\right)\)

Vận tốc trung bình của ô tô :

\(v_{tb}=\dfrac{v_1t_1+v_2t_2}{t_1+t_2}=\dfrac{54.3+36.2}{3+2}=46,8\left(km/h\right)\)

\(T=4x^2+x-9\)

\(\Leftrightarrow T=4\left(x^2+\dfrac{1}{4}x+\dfrac{1}{64}\right)-\dfrac{1}{16}-9\)

\(\Leftrightarrow T=4\left(x+\dfrac{1}{8}\right)^2-\dfrac{145}{16}\ge-\dfrac{145}{16},\forall x\in R\)

Dấu "=" xảy ra khi \(x+\dfrac{1}{8}=0\Leftrightarrow x=-\dfrac{1}{8}\)

Vậy \(GTNN\left(T\right)=-\dfrac{145}{16}\left(tại.x=-\dfrac{1}{8}\right)\)

Chọn gốc thế năng ở vị trí mặt đất 

a) \(W_t\left(A\right)=mgh=2.10.3=60\left(J\right)\)

b) \(W_t\left(B\right)=mgh=2.10.3=60\left(J\right)\)

c) \(W_t\left(C\right)=mgh=1.10.3=30\left(J\right)\)

d) \(W_t\left(D\right)=mgh=3.10.2=60\left(J\right)\)

\(s=3480\left(m\right)=3,48\left(km\right)\)

\(v=20\left(km/h\right)\)

Thời gian đi là :

\(t=\dfrac{s}{v}=\dfrac{3,48}{20}=0,174\left(h\right)=626,4\left(s\right)\)

Thời gian đi được nửa quãng đường đầu :

\(t_1=\dfrac{s_1}{v_1}=\dfrac{2}{20}=0,1\left(h\right)\)

Thời gian đi được nửa quãng đường sau :

\(t_2=\dfrac{s_2}{v_2}=\dfrac{2}{10}=0,2\left(h\right)\)

Tốc độ trung bình cả quãng đường từ nhà đến trường :

\(v_{tb}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{2+2}{0,1+0,2}=13,33\left(km/h\right)\)