Question

Answer 1

a: ĐKXĐ: x>=0

Ta có: \(3\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}=\left(3\sqrt{3x}-3\sqrt{3x}\right)-4\sqrt{3x}+27=-4\sqrt{3x}+27\)

b: ĐKXĐ: x>=0

Ta có: \(3\cdot\sqrt{2x}-5\cdot\sqrt{8x}+7\cdot\sqrt{18x}+28\)

\(=3\cdot\sqrt{2x}-5\cdot2\cdot\sqrt{2x}+7\cdot3\cdot\sqrt{2x}+28\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

c: ĐKXĐ: x>=-1/2

Ta có: \(\sqrt{8x+4}-\sqrt{32x+16}-2\cdot\sqrt{18x+9}\)

\(=2\cdot\sqrt{2x+1}-4\cdot\sqrt{2x+1}-2\cdot3\sqrt{2x+1}\)

\(=-2\sqrt{2x+1}-6\cdot\sqrt{2x+1}=-8\cdot\sqrt{2x+1}\)