Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

\(1,=\left|\sqrt{7}-4\right|-2\sqrt{7}=4-\sqrt{7}-2\sqrt{7}=4-3\sqrt{7}\\ 2,\\ a,P=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ P=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\\ b,P>3\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

\(BC=\sqrt{AB^2+AC^2}=\sqrt{3,5^2+4,2^2}=\dfrac{7\sqrt{61}}{10}cm\left(Pyatgo\right)\)

\(\sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{3,5}{\dfrac{7\sqrt{61}}{10}}\simeq39^0\)

\(\Rightarrow\widehat{B}=\widehat{A}-\widehat{C}=90^0-39^0=51^0\)

a) \(D=4\sqrt{\dfrac{1}{3}}+5\sqrt{12}-6\sqrt{27}\)

\(=\dfrac{4}{9}\sqrt{3}+5.2\sqrt{3}-6.3\sqrt{3}\)

\(=\dfrac{4}{9}\sqrt{3}+10\sqrt{3}-18\sqrt{3}\)

\(=-\dfrac{68}{9}\sqrt{3}\)

b) \(E=\dfrac{2}{\sqrt{3}-1}-\sqrt{4-2\sqrt{3}}\)

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

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

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

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

c) \(F=\dfrac{\sqrt{15}-\sqrt{10}}{\sqrt{3}-\sqrt{2}}+\dfrac{3}{2-\sqrt{5}}\)

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

\(=\sqrt{5}-6-3\sqrt{5}=-2\sqrt{5}-6\)