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a) ĐKXĐ \(\sqrt{x}\ne2\Leftrightarrow x\ne4\)
b) \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{1}{\sqrt{x}-2}\)
\(A=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
c) x = 6 + \(4\sqrt{2}\) = \(\left(2+\sqrt{2}\right)^2\)
=> A = \(\frac{\sqrt{\left(2+\sqrt{2}\right)^2}-4}{\sqrt{\left(2+\sqrt{2}\right)^2}-2}=\frac{\sqrt{2}-2}{\sqrt{2}}\)
2.
a) đkxđ: \(x\ne4;x\ne9\)
A=\(\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
=\(\left(\frac{x-\sqrt{x}-6+x\sqrt{x}-9\sqrt{x}-2x+18-x\sqrt{x}+2x-4\sqrt{x}-8}{\left(x-4\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(\frac{x-14\sqrt{x}-5}{\left(x-4\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b) A = -2/5
(k biết là do đề sai hay mình sai chứ đến đây nản quá! bạn làm nốt nhé!)
\(M=\left(\frac{\sqrt{t^2-a^2}+\sqrt{t^2+a^2}}{\sqrt{t^2-a^2}-\sqrt{t^2+a^2}}\right)^4\)
DO \(t=a\sqrt{\frac{x^2+1}{2x}}\)
=> \(t^2=a^2.\frac{x^2+1}{2x}\)
=> \(\sqrt{t^2-a^2=}\sqrt{a^2.\frac{x^2+1}{2x}-a^2}=\sqrt{a^2\left(\frac{x^2+1-2x}{2x}\right)}\)
= \(a\sqrt{\frac{\left(x-1\right)^2}{2x}}\)
TƯƠNG TỰ : \(\sqrt{t^2+a^2}=a\sqrt{\frac{\left(x+1\right)^2}{2x}}\)
=> M = \(\left(\frac{\sqrt{t^2-a^2}+\sqrt{t^2+a^2}}{\sqrt{t^2-a^2}-\sqrt{t^2+a^2}}\right)^4\)
= \(\left(\frac{a\left(\sqrt{\frac{\left(x+1\right)^2}{2x}}+\sqrt{\frac{\left(x-1\right)^2}{2x}}\right)}{a\left(\sqrt{\frac{\left(x-1\right)^2}{2x}}-\sqrt{\frac{\left(x+1\right)^2}{2x}}\right)}\right)^4\)
= \(\left(\frac{\sqrt{\frac{1}{2x}}.\left(x+1+x-1\right)}{\sqrt{\frac{1}{2x}}.\left(x-1-x-1\right)}\right)^4\)
( DO X+1>X-1>0)
= \(\left(\frac{2x}{-2}\right)^4\)
= \(x^4\)
= \(2012^4\)
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