ĐK: \(x>0;x\neq 4\)

\(a,P=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{4x}\\ P=\dfrac{2\sqrt{x}}{4x}=\dfrac{1}{2\sqrt{x}}\\ b,P>1\Leftrightarrow\dfrac{1}{2\sqrt{x}}-1>0\Leftrightarrow\dfrac{1-2\sqrt{x}}{2\sqrt{x}}>0\\ \Leftrightarrow1-2\sqrt{x}>0\left(2\sqrt{x}>0\right)\\ \Leftrightarrow\sqrt{x}< \dfrac{1}{2}\Leftrightarrow0< x< \dfrac{1}{4}\)

\(a,=5x-10+2x+6=7x-4\\ b,=x^2+2x+1-x^2+3x+10=5x+11\\ c,=x^2-49-x^2+1=-48\\ d,\text{Đề có sai ko vậy?}\)

\(b,=\dfrac{7}{2}\left(18\dfrac{2}{5}-24\dfrac{2}{5}\right)=\dfrac{7}{2}\left(-6\right)=-21\\ c,=\dfrac{3}{5}\cdot\dfrac{1}{3}+4\cdot\dfrac{1}{4}-1=\dfrac{1}{5}+1-1=\dfrac{1}{5}\)

\(5^{27}=\left(5^3\right)^9=125^9< 128^9=\left(2^7\right)^9=2^{63}\left(1\right)\\ 2^{63}=\left(2^9\right)^7=512^7< 625^7=\left(5^4\right)^7=5^{28}\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)

\(\left(10x^2y-25xy^3\right):10xy=x-\dfrac{5}{2}y^2\)

Áp dụng PTG, ta có: \(BC=\sqrt{AC^2+AB^2}=5\left(cm\right)\)

Áp dụng HTL, ta có:

\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{3\cdot4}{5}=2,4\left(cm\right)\\ HB=\dfrac{AB^2}{BC}=\dfrac{3^2}{5}=1,8\left(cm\right)\)

Vì AM là trung tuyến ứng cạnh huyền BC nên \(AM=\dfrac{1}{2}BC=2,5\left(cm\right)\)

\(=3\sqrt{5}+6\sqrt{5}\cdot\sqrt{5}-3\sqrt{5}=3\sqrt{5}+30-3\sqrt{5}=30\)

\(=3\sqrt{5}+6\sqrt{5}\cdot\sqrt{5}-3\sqrt{5}=3\sqrt{5}+30-3\sqrt{5}=30\)

\(a,=4\cdot5+14:7=20+2=22\\ b,PT\Leftrightarrow\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\1-2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ c,VT=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right]\\ =a-\sqrt{ab}+b-\sqrt{ab}=a-2\sqrt{ab}+b=\left(\sqrt{a}-\sqrt{b}\right)^2=VP\)

Vì ABCD là hcn nên \(\widehat{BAC}=\widehat{BCD}=90^0\)

\(\Rightarrow\widehat{BAC}+\widehat{BCD}=180^0\)

Do đó ABCD nội tiếp hay A,B,C,D cùng thuộc 1 đt

Ta lại có ABCD là hcn nên \(AO=BO=CO=DO\) với \(\left\{O\right\}=AC\cap BD\)

Do đó O là tâm của \(\left(ABCD\right)\)

Áp dụng PTG, ta có: \(BD=\sqrt{AB^2+BC^2}=13\left(cm\right)\)

Mà OA là trung tuyến ứng cạnh huyền BD nên \(OA=\dfrac{1}{2}BD=6,5\left(cm\right)\)

Vậy \(R\left(ABCD\right)=6,5\left(cm\right)\)