Bài 1:
a: \(M=\left(5\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
\(=5\cdot\sqrt{20}-3\cdot\sqrt{45}+2\cdot\sqrt5\)
\(=10\sqrt5-9\sqrt5+2\sqrt5=3\sqrt5\)
b: \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8=\sqrt{64}>\sqrt{63}=3\sqrt7\)
Bài 2:
a: Thay x=0,25 vào A, ta được:
\(A=\frac{2+\sqrt{0,25}}{\sqrt{0,25}}=\frac{2+0,5}{0,5}=\frac{2.5}{0.5}=5\)
b: \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c: \(\frac{A}{B}\ge\frac32\)
=>\(\frac{\sqrt{x}+2}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\ge\frac32\)
=>\(\frac{\sqrt{x}+1}{\sqrt{x}}\ge\frac32\)
=>\(1+\frac{1}{\sqrt{x}}\ge1+\frac12\)
=>\(\frac{1}{\sqrt{x}}\ge\frac12\)
=>\(\sqrt{x}<2\)
=>0<x<4
