Oh yeah! Mk chắc chắn lm đc câu 1!

Câu 2:

a) P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

ĐK : \(x\ge0;x\ne4\)

Ta có: P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

= \(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\)

\(\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{x+2\sqrt{x}+\sqrt{x}+3+2-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Để P = 2 <=> \(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)

=> \(3\sqrt{x}=2\sqrt{x}+4\)

<=> \(\sqrt{x}=4\) <=> \(x=16\left(TM\right)\)

Vì 3-2=1 và 3-1=2 nên 1+2=3

Ta có: A = \(\sqrt{\left(\sqrt{6}-2\sqrt{2}\right)^2}-\sqrt{24-12\sqrt{3}}\)

= \(\left|\sqrt{6}-2\sqrt{2}\right|\) \(-\sqrt{18-2.6\sqrt{3}+6}\)

= \(2\sqrt{2}-\sqrt{6}-\sqrt{\left(\sqrt{18}-\sqrt{6}\right)^2}\)

= \(2\sqrt{2}-\sqrt{6}-\sqrt{18}+\sqrt{6}\)

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