a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)

=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)

12\(\sqrt{3}\)

 \(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)

\(=2\sqrt{3}.6\sqrt{3}=36\)

\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)

\(=11-4\sqrt{6}\)

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

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

a)\(\sqrt{3\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{2}+2\sqrt{3}}\)

= \(\sqrt{18-12}\)

= \(\sqrt{6}\)

b) \(\sqrt{2+2\sqrt{2-\sqrt{2}}}.\sqrt{2-2\sqrt{2-\sqrt{2}}}\)

= \(\sqrt{4-4\left(\sqrt{2-\sqrt{2}}\right)^2}\)

= \(\sqrt{4-4.\left(2-4\sqrt{2}+2\right)}\)

= \(\sqrt{4-8+16\sqrt{2}-8}\)

= \(\sqrt{-12+16\sqrt{2}}\)

c) 

\(\left(\sqrt{2}-\sqrt{7}\right).\sqrt{9+2\sqrt{14}}\)

= \(\left(\sqrt{2}-\sqrt{7}\right).\left(2+2\sqrt{7}.\sqrt{2}+7\right)\)

= \(\left(\sqrt{2}-\sqrt{7}\right).\left(\sqrt{2}+\sqrt{7}\right)^2\)

= \(\left(4-7\right).\left(\sqrt{2}+\sqrt{7}\right)\)

= \(-3.\left(\sqrt{2}+\sqrt{7}\right)\)

a) \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8\)

b)\(\sqrt{9\left(3-a\right)^2}=3\left|3-a\right|=3\left(a-3\right)\)(vì a > 3)

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)

\(=\sqrt{\left(\sqrt{9}\right)^2}-\sqrt{\left(\sqrt{17}\right)^2}\)

\(\sqrt{9\left(3-a\right)^2}\)

\(=\sqrt{3^2\left(3-a\right)^2}\)

\(=3\left(3-a\right)\)

\(=3-3a\)

a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

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

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

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

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

Câu 2b đề là tìm x chứ nhỉ???

b) \(\sqrt{x^2-4}+\sqrt{x-2}=0\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x-2}\ge0\end{matrix}\right.\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\pm2\\x=2\end{matrix}\right.\) <=> x = 2

Vậy x = 2

bài 2 câu b) đề sai rồi bạn

còn bài 1 câu b) mình cảm thấy sai sai

\(=6\sqrt{2}+\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\cdot\sqrt{\dfrac{8}{3}}+6\sqrt{3}\)

\(=6\sqrt{2}+\dfrac{4}{3}\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=6\sqrt{2}+3\sqrt{6}+\dfrac{22}{3}\sqrt{3}\)