A

Câu 31. D

Câu 32. D

unhealthily

Câu 29. C

Câu 30. A

Ta có:

\(A=\sqrt{11+\sqrt{96}}\)

\(=\sqrt{11+\sqrt{4.24}}\)

\(=\sqrt{11+\sqrt{2^2.24}}\)

\(=\sqrt{11+2\sqrt{24}}\)

\(=\sqrt{8+3+2\sqrt{8}.\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{8}+\sqrt{3}\right)^2}\)

\(=\left|\sqrt{8}+\sqrt{3}\right|\)

\(=\sqrt{8}+\sqrt{3}\) (vì \(\sqrt{8}+\sqrt{3}>0\))

\(=\sqrt{4.2}+\sqrt{3}\)

\(=\sqrt{2^2.2}+\sqrt{3}\)

\(=2\sqrt{2}+\sqrt{3}\)

\(=\sqrt{2}+\sqrt{2}+\sqrt{3}\)

\(B=\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)

\(=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}\)

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

\(=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{3+2\sqrt{2}-3}\)

\(=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}\)

\(=1+\sqrt{2}+\sqrt{3}\)

Vì \(2>1\Leftrightarrow\sqrt{2}>\sqrt{1}\Leftrightarrow\sqrt{2}>1\Leftrightarrow\sqrt{2}+\sqrt{2}+\sqrt{3}>1+\sqrt{2}+\sqrt{3}\Leftrightarrow A>B\)

\(A=\left(x-2\right)^2+3\ge3\)

Dấu bằng xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(MinA=3\Leftrightarrow x=2\)

\(B=\left(x-4\right)^2-16\ge-16\)

Dấu bằng xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy \(MinB=-16\Leftrightarrow x=4\)

\(C=-2\left(x-2\right)^2-7\le-7\)

Dấu bằng xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(MaxC=-7\Leftrightarrow x=2\)

\(D=-\left(x-3\right)^2-\left(y-1\right)^2+2\le2\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy \(MaxD=2\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

\(E=\left(x^2+1\right)^2+4\ge\left(0+1\right)^2+4=1+4=5\)

Dấu bằng xảy ra \(\Leftrightarrow x=0\)

Vậy \(MinE=5\Leftrightarrow x=0\)

Điều kiện xác định: \(x\ge3\)

\(3\sqrt{9x-27}-\dfrac{2}{5}\sqrt{25x-75}=\dfrac{1}{2}\sqrt{4x-12}\)

\(\Leftrightarrow3\sqrt{9\left(x-3\right)}-\dfrac{2}{5}\sqrt{25\left(x-3\right)}=\dfrac{1}{2}\sqrt{4\left(x-3\right)}\)

\(\Leftrightarrow3.3\sqrt{x-3}-\dfrac{2}{5}.5\sqrt{x-3}=\dfrac{1}{2}.2\sqrt{x-3}\)

\(\Leftrightarrow9\sqrt{x-3}-2\sqrt{x-3}=\sqrt{x-3}\)

\(\Leftrightarrow6\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\) (thỏa mãn)

Vậy phương trình có tập nghiệm là: \(S=\left\{3\right\}.\)

\(1\in A,\) \(6\notin B,\) \(4\notin A\)

Điều kiện xác định: \(n+2\ne0\Leftrightarrow n\ne-2\)

\(\dfrac{n+5}{n+2}=\dfrac{\left(n+2\right)+3}{n+2}=1+\dfrac{3}{n+2}\)

Yêu cầu bài toán \(\Leftrightarrow3⋮\left(n+2\right)\Leftrightarrow\left(n+2\right)\inƯ\left(3\right)\)

Mà \(Ư\left(3\right)=\left\{-3;-1;1;3\right\}.\)

\(\Rightarrow\left[{}\begin{matrix}n+2=-3\\n+2=-1\\n+2=1\\n+2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-5\\n=-3\\n=-1\\n=1\end{matrix}\right.\)

So với điều kiện xác định thấy thỏa mãn.

Vậy \(n\in\left\{-5;-3;-1;1\right\}\) thỏa mãn yêu cầu bài toán.

a)x^2+4x+4=(x+2)^2                       d)x^3+12x^2+48x+64=(x+4)^3

b)x^2-8x+16=(x-4)^2                          e)x^3-6x+12x-8=(x-2)^3

c) (x+5)(x-5)=x^2-25                         f) (x+2)(x^2-2x+4)=x^3+8