\(\widehat{A}+\widehat{B}+\widehat{C}=180^o=>2.\widehat{C}+2.\widehat{C}+\widehat{C}=180^o\\ =>5.\widehat{C}=180^0\\ =>\widehat{C}=\dfrac{180^o}{5}=36^o\\ =>\widehat{A}=\widehat{B}=2.36^o=72^o\)

\(\left(2x-1\right)^3+8=0\\ < =>\left(2x-1\right)^3=-8=\left(-2\right)^3\\ =>2x-1=-2\\ < =>x=-\dfrac{1}{2}\)

\(x^3-3x^2+3x-2=0\\ < =>x^3-3.x^2.1+3.x.1^2-1^3-1=0\\ < =>\left(x-1\right)^3=1=1^3\\ =>x-1=1\\ < =>x=2\)

\(ĐK:x\in Z;x\ne1\\ 5⋮\left(x-1\right)=>x-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\\ =>x\in\left\{0;2;-4;6\right\}\left(TMDK\right)\)

\(\left(x+1\right)^4=\left(2x\right)^4=\left(\pm2x\right)^4\\ =>\left[{}\begin{matrix}x+1=2x\\x+1=-2x\end{matrix}\right.< =>\left[{}\begin{matrix}2x-x=1\\x+2x=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\left(2x-1\right)^5=x^5\\ =>2x-1=x\\ =>2x-x=1\\ =>x=1\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\left(ĐK:4x^2-9>=0;2x+3>=0\right)\\ < =>4x^2-9=4\left(2x+3\right)\\ < =>\left(2x+3\right)\left(2x-3\right)-4\left(2x+3\right)=0\\ < =>\left(2x+3\right)\left(2x-3-4\right)=0\\ < =>\left(2x+3\right)\left(2x-7\right)=0\\ =>\left[{}\begin{matrix}2x+3=0\\2x-7=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(TMDK\right)\\x=\dfrac{7}{2}\left(TMDK\right)\end{matrix}\right.=>S=\left\{-\dfrac{3}{2};\dfrac{7}{2}\right\}\)

\(M=\left(1-\dfrac{4\sqrt{x}}{x-1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-2\sqrt{x}}{x-1}\left(x>0;x\ne\left\{1;4\right\}\right)\\ =\left(1-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{x-1}{x-2\sqrt{x}}\\ =\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

\(M=\dfrac{1}{2}< =>\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{1}{2}\\ < =>2\left(\sqrt{x}-3\right)=\sqrt{x}-2\\ < =>2\sqrt{x}-6=\sqrt{x}-2\\ < =>\sqrt{x}=4\\ < =>x=16\left(TMDK\right)\)

\(2.16.8=2^1.2^4.2^3=2^{1+4+3}=2^8\)

\(25.5.125=5^2.5^1.5^3=5^{2+1+3}=5^6\)

\(\dfrac{2}{3}.\dfrac{4}{9}.\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^1.\left(\dfrac{2}{3}\right)^2.\left(\dfrac{2}{3}\right)^3=\left(\dfrac{2}{3}\right)^{1+2+3}=\left(\dfrac{2}{3}\right)^6\)

\(A=\left\{x\in N|x⋮2,0< x< 12\right\}\)

\(=>\left[{}\begin{matrix}\sqrt{x}-4=\sqrt{20}\\\sqrt{x}-4=-\sqrt{20}\end{matrix}\right.< =>\left[{}\begin{matrix}\sqrt{x}=4+2\sqrt{5}\\\sqrt{x}=4-2\sqrt{5}< 0\left(KTM\right)\end{matrix}\right.=>x=\left(4+2\sqrt{5}\right)^2=36+16\sqrt{5}\left(TMDK\right)\)

\(ĐK:\left\{{}\begin{matrix}x>=0\\\sqrt{x}-2\ne0\\x\sqrt{x}-8\ne0\\\sqrt{x}+2\ne0\end{matrix}\right.< =>x>=0;x\ne4\)

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x\left(\sqrt{x}+2\right)}{x\sqrt{x}-8}\right).\left(\dfrac{x}{\sqrt{x}+2}+2\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right).\dfrac{x+2\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\dfrac{x+2\sqrt{x}+4}{\sqrt{x}+2}\\ =\dfrac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}-2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{4\sqrt{x}}{x-4}\)

\(A=\dfrac{1}{2}< =>\dfrac{4\sqrt{x}}{x-4}=\dfrac{1}{2}\\ < =>x-4=8\sqrt{x}< =>x-8\sqrt{x}-4=0\\ < =>\left(\sqrt{x}-4\right)^2-20=0\\ =>\left[{}\begin{matrix}\sqrt{x}-4=\sqrt{20}\\\sqrt{x}-4=-\sqrt{20}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\left(4+2\sqrt{5}\right)^2=36+16\sqrt{5}\left(TMDK\right)\\x=\left(4-2\sqrt{5}\right)^2=36-16\sqrt{5}\left(TMDK\right)\end{matrix}\right.\)