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Решите задачу:

$1)\frac{2+\sqrt{a}}{a+2\sqrt{a}+1 }-\frac{\sqrt{a}-2}{a-1}=\frac{2+\sqrt{a}}{(\sqrt{a})^{2}+2\sqrt{a}+1^{2}}-\frac{\sqrt{a}-2 }{(\sqrt{a})^{2}-1^{2}}=\frac{2+\sqrt{a}}{(\sqrt{a}+1)^{2}}-\frac{\sqrt{a}-2}{(\sqrt{a}+1)(\sqrt{a}-1)}=\frac{2\sqrt{a}-2+a-\sqrt{a}-a+2\sqrt{a}-\sqrt{a}+2}{(\sqrt{a}+1)^{2}(\sqrt{a}-1)}=\frac{2\sqrt{a} }{(\sqrt{a}+1)^{2}(\sqrt{a}-1)}$

$2)\frac{a\sqrt{a}+a-\sqrt{a}-1}{\sqrt{a}}=\frac{(a\sqrt{a}+a)-(\sqrt{a}+1)}{\sqrt{a}}=\frac{a(\sqrt{a}+1)-(\sqrt{a} +1)}{\sqrt{a}}=\frac{(\sqrt{a}+1)(a-1)}{\sqrt{a}}=\frac{(\sqrt{a}+1)(\sqrt{a}+1)(\sqrt{a}-1)}{\sqrt{a}}=\frac{(\sqrt{a}+1)^{2}(\sqrt{a}-1)}{\sqrt{a}}$

$3)\frac{2\sqrt{a}}{(\sqrt{a}+1)^{2}(\sqrt{a}-1)}*\frac{(\sqrt{a}+1)^{2}(\sqrt{a}-1)}{\sqrt{a}}=2$