c, \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

<=> \(C=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)

<=> \(C=-5\sqrt{3}:\sqrt{3}=-5\)

e. \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

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

\(=6+2\sqrt{9-5}\)

\(=6+4=10\)

b. \(\left(\sqrt{3}+2\right)^2-\sqrt{75}\)

\(=3+4\sqrt{3}+4-5\sqrt{3}\)

\(=7-\sqrt{3}\)

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

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

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

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

f. \(\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

\(=2\sqrt{3}\)

c: Ta có: \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-5\cdot3\sqrt{3}+4\cdot2\sqrt{3}\right):\sqrt{3}\)

\(=2-15+8=-5\)

d: Ta có: \(D=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

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

\(=6+2\cdot2=10\)

\(\Leftrightarrow7\left(2x-1\right)-15x=-3x\)

=>14x-7-15x+3x=0

=>2x=7

hay x=7/2(nhận)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{7}{3x}-\dfrac{5}{2x-1}=\dfrac{1}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}+\dfrac{5}{1-2x}=\dfrac{1}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}=\dfrac{1}{1-2x}-\dfrac{5}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}=\dfrac{-4}{1-2x}\)

\(\Rightarrow-4.3x=7\left(1-2x\right)\)

\(\Rightarrow-12x=7-14x\)

\(\Rightarrow-12x+14x=7\)

\(\Rightarrow2x=7\)

\(\Rightarrow x=\dfrac{7}{2}\left(tm\right)\)

PT⇔7/3x=5/2x-1 - 1/2x-1

     <=>7/3x=4/2x-1

     <=>7(2x-1)=4.3x

      <=>14x-7=12x

       <=>14x-12x=7

       <=>2x=7

       <=>x=7/2

Vậy tập nghiệm của pt là S={7/2}

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