Bài 27.

a) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)

b) \(8-12x+6x^2-x^3=\left(2-x\right)^3\)

Bài 28.

a) \(x^3+12x^2+48x+64=\left(x+4\right)^3\)

Tại \(x=6\) ta có \(\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

b) \(x^3-6x^2+12x-8=\left(x-2\right)^3\)

Tại \(x=22\) ta có \(\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

\(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{3}{1-\sqrt{x}}+\dfrac{x+8}{x+\sqrt{x}-2}\\ =\dfrac{1}{\sqrt{x}+2}-\dfrac{3}{\sqrt{x}-1}+\dfrac{x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)-3\left(\sqrt{x}+2\right)+x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(A=1-2+3-4+5-6+7-...-100+101\\ =\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(99-100\right)+101\\ =\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)+101\\ =\left(-1\right)\times50+101\\ =-50+101\\ =51\)

ĐK: \(x\ge-\dfrac{8}{3}\)

\(\sqrt{x^2+4}=\sqrt{3x+8}\\ \Rightarrow x^2+4=3x+8\\ \Rightarrow x^2-3x-4=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\left(TM\right)\)

\(6\sqrt{\dfrac{3}{4}}-\dfrac{2}{\sqrt{3}-2}-\sqrt{4-2\sqrt{3}}\\ =\dfrac{6\sqrt{3}}{\sqrt{4}}-\dfrac{2\left(\sqrt{3}+2\right)}{3-2^2}-\sqrt{3-2\sqrt{3}+1}\\ =\dfrac{6\sqrt{3}}{2}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\sqrt{\left(\sqrt{3}-1\right)^2}\\ =3\sqrt{3}+2\left(\sqrt{3}+2\right)-\left(\sqrt{3}-1\right)\\ =4\sqrt{3}+5\)

a) \(A=\left\{1233;1323;1332;2133;2313;2331;3123;3132;3213;3231;3312;3321\right\}\)

b) 1233; 1332; 2133; 2331; 3132; 3231

\(\dfrac{81}{108}=\dfrac{27\times3}{27\times4}=\dfrac{3}{4}\\ \dfrac{306}{561}=\dfrac{51\times6}{51\times11}=\dfrac{6}{11}\\ \dfrac{185}{333}=\dfrac{37\times5}{37\times9}=\dfrac{5}{9}\\ \dfrac{90}{8}=\dfrac{2\times45}{2\times4}=\dfrac{45}{4}\)