1: \(\sqrt{64}+\sqrt[3]{-125}\)
\(=8+\left(-5\right)\)
=8-5
=3
2:
ĐKXĐ: x>=2
\(5\sqrt{4x-8}-\dfrac{1}{3}\cdot\sqrt{9x-18}=18\)
=>\(5\cdot2\sqrt{x-2}-\dfrac{1}{3}\cdot3\sqrt{x-2}=18\)
=>\(9\sqrt{x-2}=18\)
=>\(\sqrt{x-2}=2\)
=>x-2=22=4
=>x=6(nhận)
3:
a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}}\right):\dfrac{2x-1}{3x}\)
\(=\dfrac{x+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{3x}{2x-1}\)
\(=\dfrac{x+x-1}{\sqrt{x}+1}\cdot\dfrac{3\sqrt{x}}{2x-1}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
b: Để A là số nguyên thì \(3\sqrt{x}⋮\sqrt{x}+1\)
=>\(3\sqrt{x}+3-3⋮\sqrt{x}+1\)
=>\(-3⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;3\right\}\)
=>\(\sqrt{x}\in\left\{0;2\right\}\)
=>\(x\in\left\{0;4\right\}\)
Kết hợp ĐKXĐ, ta được: x=4
