Bài 2:
1: Khi x=16 thì \(A=\dfrac{\sqrt{16}-2}{\sqrt{16}-3}=\dfrac{4-2}{4-3}=\dfrac{2}{1}=2\)
2: \(B=\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}-\dfrac{x}{4-x}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2+\sqrt{x}-2+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
P=A*B
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
3: Để P>=0 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-3}>=0\)
mà \(\sqrt{x}>=0\)
nên \(\sqrt{x}-3>0\)
=>\(\sqrt{x}>3\)
=>x>9
Bài 1:
1:
a: \(A=\sqrt{8}+2\sqrt{18}-3\sqrt{32}\)
\(=2\sqrt{2}+2\cdot3\sqrt{2}-3\cdot4\sqrt{2}\)
\(=2\sqrt{2}+6\sqrt{2}-12\sqrt{2}\)
\(=-4\sqrt{2}\)
b: \(B=\dfrac{4}{\sqrt{5}-1}+\sqrt{\left(\sqrt{5}-1\right)^2}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{4\left(\sqrt{5}+1\right)}{4}+\left|\sqrt{5}-1\right|-2\sqrt{5}\)
\(=\sqrt{5}+1+\sqrt{5}-1-2\sqrt{5}\)
=0
2:
ĐKXĐ: x>=1
\(\sqrt{16x-16}-\sqrt{4x-4}+\sqrt{x-1}=2\)
=>\(4\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=2\)
=>\(3\sqrt{x-1}=2\)
=>\(\sqrt{x-1}=\dfrac{2}{3}\)
=>\(x-1=\dfrac{4}{9}\)
=>\(x=\dfrac{4}{9}+1=\dfrac{13}{9}\left(nhận\right)\)
