\(2024+202,4+20,24\cdot43+0,2024\cdot3700\)

\(=2024+\frac{2024}{10}+\frac{2024\cdot43}{100}+\frac{2024.3700}{10000}\)

\(=2024\left(1+\frac{1}{10}+\frac{43}{100}+\frac{37}{100}\right)\)

\(=2024\left(1+\frac{1}{10}+\frac{80}{100}\right)=2024\left(\frac{100+10+80}{100}\right)\)

\(=2024\cdot\frac{190}{100}=1012\cdot\frac{19}{5}=\frac{19228}{5}=3845,6\)

Chiều cao hình tam giác :

\(20:\frac53=20x\frac35=12\left(\operatorname{cm}\right)\)

Diện tích hình tam giác là :

\(\frac12x20x12=120\left(\operatorname{cm}^2\right)\)

Đáp số \(120\left(\operatorname{cm}^2\right)\)

Sửa lại đề bài \(S=\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}+\frac{7}{33.43}+\cdots+\frac{7}{53.63}\)

Theo đề bài ta có phân số quy luật sau

\(\frac{7}{\left(10n-7\right)\left(10n+3\right)}\left(n=1\rarr6\right)\)

\(=\frac{7}{10}\left(\frac{1}{10n-7}-\frac{1}{10n+3}\right)\)

\(\rArr S=\frac{7}{10}\left(\frac13-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+\frac{1}{33}-\frac{1}{43}+\frac{1}{43}-\frac{1}{53}+\frac{1}{53}-\frac{1}{63}\right)\) \(\rArr S=\frac{7}{10}\left(\frac13-\frac{1}{63}\right)=\frac{7}{10}.\frac{20}{63}=\frac29\)

Vậy \(S=\frac29\)

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