\(A=\dfrac{-x+5+2}{x-5}=-1+\dfrac{2}{x-5}\)

Để A đạt min thì \(\left\{{}\begin{matrix}A\in Z\\x\in Z\end{matrix}\right.\)

\(\Rightarrow x-5\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Rightarrow x\in\left\{3;4;6;7\right\}\)

Với \(x=3\Rightarrow A=-2\)

Với \(x=4\Rightarrow A=-3\)

Với \(x=6\Rightarrow A=1\)

Với \(x=7\Rightarrow A=0\)

Vậy \(A_{min}=-3\Rightarrow x=4\)

đừng lấy nó ko ra min đâu

\(a,\left\{{}\begin{matrix}AE=AB\\\widehat{BAO}=\widehat{EAO}\\AO\text{ chung}\end{matrix}\right.\Rightarrow\Delta OAE=\Delta OAB\left(c.g.c\right)\\ \Rightarrow OE=OB\\ b,\text{Gọi }\left\{H\right\}=OA\cap BE\\ \Rightarrow\left\{{}\begin{matrix}AE=AB\\\widehat{BAH}=\widehat{EAH}\\AH\text{ chung}\end{matrix}\right.\Rightarrow\Delta HAE=\Delta HAB\left(c.g.c\right)\\ \Rightarrow\left\{{}\begin{matrix}HE=HB\\\widehat{AHE}=\widehat{AHB}\end{matrix}\right.\\ \text{Mà }\widehat{AHE}+\widehat{AHB}=180^0\\ \Rightarrow\widehat{AHE}=90^0\\ \Rightarrow AH\bot BE\\ \text{Mà }HE=HB\Rightarrow AH\text{ hay }AO\text{ là trung trực }BE\)

\(c,\widehat{ABO}=\widehat{AEO}\Rightarrow180^0-\widehat{ABO}=180^0-\widehat{AEO}\\ \Rightarrow\widehat{OBD}=\widehat{OEC}\\ \text{Mà }\widehat{BOD}=\widehat{EOC}\left(đđ\right);OB=OE\\ \Rightarrow\Delta OBD=\Delta OEC\left(g.c.g\right)\\ \Rightarrow BD=EC\\ \Rightarrow AB+BD=EC+AE\\ \Rightarrow AC=BD\)

\(a,f\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^2+4=\dfrac{1}{4}+4=\dfrac{17}{4}\\ f\left(5\right)=5^2+4=25+4=29\\ b,f\left(x\right)=10\Rightarrow x^2+4=10\Rightarrow x^2=6\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\)

\(1,\\ a,x=\dfrac{2\cdot3}{6}=1\\ b,x=\dfrac{-12\cdot5}{8}=-7,5\left(ktm\right)\\ c,\Rightarrow2x^2=32\Rightarrow x^2=16\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ 2,\\ a,\Rightarrow6x+6=6\Rightarrow x=0\\ b,\Rightarrow2-2x=4\Rightarrow2x=-2\Rightarrow x=-1\\ c,\Rightarrow-2x=18\Rightarrow x=-9\\ 3,\\ a,\Rightarrow36-4x=-60\Rightarrow4x=96\Rightarrow x=24\\ b,\Rightarrow\left(x+1\right)^2=9\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\\ c,\Rightarrow\left(x-1\right)^2=16\Rightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ 4,\\ a,\Rightarrow4x=6x-3\Rightarrow2x=3\Rightarrow x=\dfrac{3}{2}\left(ktm\right)\\ b,\Rightarrow9x+3=8x-4\Rightarrow x=-7\\ 5,\\ a,\Rightarrow12x+4=7x+14\Rightarrow5x=10\Rightarrow x=2\\ b,\Rightarrow6x^2-6x=6x^2+6x\Rightarrow x=0\)

Gọi số giấy vụn 6A,6B,6C lần lượt là a,b,c>0;kg

Áp dụng tc dtsbn:

\(\dfrac{a}{43}=\dfrac{b}{44}=\dfrac{c}{45}=\dfrac{a+b+c}{43+44+45}=\dfrac{132}{132}=1\\ \Rightarrow\left\{{}\begin{matrix}a=43\\b=44\\c=45\end{matrix}\right.\)

Vậy ...

\(2n^3-6=10\Rightarrow2n^3=16\Rightarrow n^3=8=2^3\Rightarrow n=2\\ 2n^2-8=10\Rightarrow2n^2=18\Rightarrow n^2=9\Rightarrow\left[{}\begin{matrix}n=3\\n=-3\end{matrix}\right.\)

Áp dụng HTL: \(AH^2=BH\cdot HC\Rightarrow BH=\dfrac{AH^2}{HC}=\dfrac{12^2}{16}=9\left(cm\right)\)

Áp dụng PTG: \(AC=\sqrt{AH^2+HC^2}=20\left(cm\right)\)

\(\Rightarrow\cos C=\dfrac{HC}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\)

\(40\left(m^2\right)=400000\left(cm^2\right)\)

Số gạch men cần dùng là \(400000:\left(20\times20\right)=1000\)