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a)
A = 15 x − 11 x + 2 x − 3 + 3 x − 2 1 − x − 2 x + 3 x + 3 = 15 x − 11 − ( 3 x − 2 ) ( x + 3 ) − ( 2 x + 3 ) ( x − 1 ) ( x − 1 ) ( x + 3 ) = − 5 x + 7 x − 2 ( x − 1 ) ( x + 3 ) = ( x − 1 ) ( − 5 x + 2 ) ( x − 1 ) ( x + 3 ) = − 5 x + 2 x + 3 A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}=\frac{15\sqrt{x}-11-(3\sqrt{x}-2)(\sqrt{x}+3)-(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{-5x+7\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{(\sqrt{x}-1)(-5\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3} A = x + 2 x − 3 1 5 x − 1 1 + 1 − x 3 x − 2 − x + 3 2 x + 3 = ( x − 1 ) ( x + 3 ) 1 5 x − 1 1 − ( 3 x − 2 ) ( x + 3 ) − ( 2 x + 3 ) ( x − 1 ) = ( x − 1 ) ( x + 3 ) − 5 x + 7 x − 2 = ( x − 1 ) ( x + 3 ) ( x − 1 ) ( − 5 x + 2 ) = x + 3 − 5 x + 2
b)
A + 5 = − 5 x + 2 x + 3 + 5 = 17 x + 3 ≤ 17 3 ⇒ A ≤ 2 3 A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{17}{\sqrt{x}+3}\leq \frac{17}{3}\Rightarrow A\leq \frac{2}{3} A + 5 = x + 3 − 5 x + 2 + 5 = x + 3 1 7 ≤ 3 1 7 ⇒ A ≤ 3 2
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