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5: 

câu lệnh: repeat 8 [ fd 100 rt 60 ]

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Câu lệnh: repeat 8 [ fd 100 rt 60 ]

A

D

a

\(3A=3^2+3^3+...+3^{101}\)

=>\(A=\dfrac{3^{101}-3}{2}\)

\(\left\{{}\begin{matrix}SO\perp BC\\SO\perp CA\end{matrix}\right.\) \(\Rightarrow SO\perp\left(ABC\right)\)

\(AA'=\dfrac{a\sqrt{3}}{2}\) (trung tuyến tam giác đều) \(\Rightarrow AO=\dfrac{2}{3}AA'=\dfrac{a\sqrt{3}}{3}\)

\(\Rightarrow M\) nằm trên đoạn thẳng OA'

Qua M kẻ đường thẳng song song BC cắt AB và AC lần lượt tại D và E

Trong mp (SAA'), qua M kẻ đường thẳng song song SO cắt SA' tại F

Trong mp (SBC), qua F kẻ đường thẳng song song BC cắt SB và SC lần lượt tại G và H

\(\Rightarrow\) Hình thang DEHG là thiết diện của (P) và chóp

\(FM||SO\Rightarrow FM\perp\left(ABC\right)\Rightarrow FM\perp ED\)

Áp dụng định lý Talet cho tam giác ABC:

\(\dfrac{DE}{BC}=\dfrac{AM}{AA'}\Rightarrow DE=\dfrac{BC.AM}{AA'}=\dfrac{a.x}{\dfrac{a\sqrt{3}}{2}}=\dfrac{2x\sqrt{3}}{3}\)

Talet tam giác SOA':

\(\dfrac{FM}{SO}=\dfrac{MA'}{OA'}\Rightarrow FM=\dfrac{SO.MA'}{OA'}=\dfrac{2a.\left(\dfrac{a\sqrt{3}}{2}-x\right)}{\dfrac{a\sqrt{3}}{6}}=6a-4\sqrt{3}x\)

Talet tam giác SBC:

\(\dfrac{GH}{BC}=\dfrac{SF}{SA'}=1-\dfrac{FA'}{SA'}=1-\dfrac{FM}{SO}=1-\dfrac{6a-4\sqrt{3}x}{2a}=\dfrac{2\sqrt{3}x-2a}{a}\)

\(\Rightarrow GH=2\sqrt{3}x-2a\)

\(S_{DEHG}=\dfrac{1}{2}\left(DE+GH\right).FM=\dfrac{1}{2}\left(\dfrac{2x\sqrt{3}}{3}+2\sqrt{3}x-2a\right)\left(6a-4\sqrt{3}x\right)\)

\(=\dfrac{1}{3}\left(4\sqrt{3}x-3a\right)\left(6a-4\sqrt{3}x\right)\le\dfrac{1}{12}\left(4\sqrt{3}x-3a+6a-4\sqrt{3}x\right)^2=\dfrac{9a^2}{12}\)

Dấu "=" xảy ra khi \(4\sqrt{3}x-3a=6a-4\sqrt{3}x\Leftrightarrow x=\dfrac{9a}{8\sqrt{3}}=\dfrac{3a\sqrt{3}}{8}\)

Ta có: (u.v)' = u'.v + u.v'

\(Q=80K^{\dfrac{1}{3}}\left(100-K\right)^{\dfrac{1}{2}}\)

\(Q'=80.\left(K^{\dfrac{1}{3}}\right)'.\left(100-K\right)^{\dfrac{1}{2}}+80.K^{\dfrac{1}{3}}.\left(\left(100-K\right)^{\dfrac{1}{2}}\right)'\)= \(80.\dfrac{1}{3}.K^{-\dfrac{2}{3}}.\left(100-K\right)^{\dfrac{1}{2}}+80.K^{\dfrac{1}{3}}.\dfrac{1}{2}.\left(100-K\right)^{-\dfrac{1}{2}}.\left(-1\right)\) = \(80.\left(\dfrac{\left(100-K\right)^{\dfrac{1}{2}}}{3K^{\dfrac{2}{3}}}-\dfrac{K^{\dfrac{1}{3}}}{2\left(100-K\right)^{\dfrac{1}{2}}}\right)\)= \(80.\left(\dfrac{2\left(100-K\right)^{\dfrac{1}{2}}\left(100-K\right)^{\dfrac{1}{2}}-3K^{\dfrac{2}{3}}K^{\dfrac{1}{3}}}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(80.\left(\dfrac{2\left(100-K\right)-3K}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(80.\left(\dfrac{200-5K}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(\dfrac{400\left(40-K\right)}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\) = \(\dfrac{200\left(40-K\right)}{3K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\).

\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)

\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)

\(\Leftrightarrow2x+8=-x+11\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Nhân chéo ta được\(-6(x+4)=3(x-11)=>-6x-24=3x-33=>6x-3x-24+33=0=>3x+9=0=>3x=-9=>x=-3\)