cho diểm B nằm giữa 2 điểm A và C;AB=2cm,BC=5cm.gọi I,M,M theo thứ tự là trung điểm của AC,AB,BC a)tính BI b)tính MN
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\(y=x^3-3x^2-9x+35\)
\(y'=3x^2-6x-9\)
\(y'=0\Leftrightarrow3x^2-6x-9=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
\(y\left(-4\right)=-41;y\left(-1\right)=40;y\left(3\right)=8;y\left(4\right)=52\)
\(\Rightarrow y_{max}=y\left(4\right)=52;y_{min}=y\left(-4\right)=-41\) trên đoạn \(\left[-4;4\right]\)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
Gọi số đó là: \(\overline{ab}\)
Theo đề bài số đó gấp 8 lần tổng các chữ số nó, vậy:
\(\overline{ab}=8\times\left(a+b\right)\\ \Rightarrow a\times10+b=8\times a+8\times b\\ \Rightarrow a\times10-8\times a=8\times b-b\\ \Rightarrow2\times a=7\times b\\ \Rightarrow\overline{ab}=27\)
Vậy số đó bằng 27