\(=2+2+2+...+2\)

Tổng trên có \(\left[\left(20-2\right):2+1\right]:2=5\) số hạng 2

Vậy tổng là \(5\times2=10\)

\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=\dfrac{a-b+c}{2-5+7}=\dfrac{a+2b-c}{2+10-7}\\ \Rightarrow A=\dfrac{2-5+7}{2+10-7}=\dfrac{4}{5}\)

Gọi số hs xuất sắc 7A,7B,7C lần lượt là a,b,c∈N*

Áp dụng tc dtsbn:

\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{2}=\dfrac{3c-b}{6-4}=\dfrac{6}{2}=3\\ \Rightarrow\left\{{}\begin{matrix}a=9\\b=12\\c=6\end{matrix}\right.\)

Vậy ...

Gọi số hs là \(x(x\in \mathbb{N^*})\)

Ta có \(3=3;4=2^2;5=5;9=3^2\)

\(\Rightarrow BCNN\left(3,4,5,9\right)=3^2\cdot2^2\cdot5=180\\ \Rightarrow BC\left(3,4,5,9\right)=B\left(180\right)=\left\{0;180;360;...;1620;1800;1980;2160;...\right\}\)

Mà \(x\in BC\left(3,4,5,9\right);1600< x< 2000\)

Vậy số hs có thể là \(1620,1800,1980\) hs

\(\left(x^3-x^2-7x+3\right):\left(x-3\right)\\ =\left(x^3-3x^2+2x^2-6x-x+3\right):\left(x-3\right)\\ =\left(x-3\right)\left(x^2+2x-1\right):\left(x-3\right)=x^2+2x-1\)

\(\left(3x^3+ax+27\right)⋮\left(x+5\right)\\ \Leftrightarrow3x^3+ax+27=\left(x+5\right)\cdot a\left(x\right)\)

Thay \(x=-5\Leftrightarrow-375-5a+27=0\\ \Leftrightarrow-5a=348\Leftrightarrow a=-\dfrac{348}{5}\)

\(a,AB=R\sqrt{2}\Rightarrow sđ\stackrel\frown{AB}=90^0\)

Do \(AP\bot MB,BQ\bot MA\Rightarrow sđ\stackrel\frown{MP}=sđ\stackrel\frown{MQ}=90^0\)

\(\Rightarrow sđ\stackrel\frown{PMQ}=180^0\Rightarrow PQ\) là đường kính

\(b,\left\{{}\begin{matrix}MA\text{//}BS\left(\bot BQ\right)\\MB\text{//}AS\left(\bot AP\right)\end{matrix}\right.\Rightarrow AMBS\) là hbh

\(c,\widehat{AQB}=\dfrac{1}{2}sđ\stackrel\frown{AB}=45^0\\ \Rightarrow\Delta HAQ\text{ vuông cân tại }A\\ \Rightarrow QA=AH\\ PS\text{//}AM\Rightarrow\widehat{APS}=\widehat{PAM}=\dfrac{1}{2}sđ\stackrel\frown{PM}=45^0\\ \Rightarrow\Delta SAP\text{ vuông cân tại }A\\ \Rightarrow AP=AS\\ \Rightarrow\Delta SAH=\Delta PAQ\left(c.g.c\right)\\ \Rightarrow SH=PQ=2R\left(const\right)\)

Bài 1:

Bài 2: