Question

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Answer 1

Câu 1:

\(a,P=\dfrac{x-5\sqrt{x}+x+6\sqrt{x}+5-5+9\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{2x+10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}-5}\\ c,P< 1\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}-5}-1< 0\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}+5}{\sqrt{x}-5}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}-5}< 0\Leftrightarrow\sqrt{x}-5< 0\left(\sqrt{x}+5>0\right)\\ \Leftrightarrow0\le x< 25\)

Câu 2:

\(A=\dfrac{2\left(3+\sqrt{5}\right)}{4+\sqrt{6+2\sqrt{5}}}+\dfrac{2\left(3-\sqrt{5}\right)}{4-\sqrt{6-2\sqrt{5}}}\\ A=\dfrac{6+2\sqrt{5}}{5+\sqrt{5}}+\dfrac{6-2\sqrt{5}}{5-\sqrt{5}}\\ A=\dfrac{\left(\sqrt{5}+1\right)^2}{\sqrt{5}\left(\sqrt{5}+1\right)}+\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}\left(\sqrt{5}-1\right)}\\ A=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{5}}=\dfrac{2\sqrt{5}}{2}=2\)

Answer 2

Câu 3:

\(B=\left(2\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\\ B=\left[\left(2\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\right]^2-8\sqrt{20+6\sqrt{3}+8}\\ B=\left(3\sqrt{3}+4\right)^2-8\sqrt{\left(3\sqrt{3}+1\right)^2}\\ B=43+24\sqrt{3}-24\sqrt{3}-8=35\)

Câu 4:

\(a,M=\dfrac{3x-3\sqrt{xy}-3x+x+\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}\cdot\dfrac{2\left(x+\sqrt{xy}+y\right)}{\left(x-1\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ M=\dfrac{2\left(x-2\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2\left(x-1\right)}=\dfrac{2\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)^2\left(x-1\right)}=\dfrac{2}{x-1}\\ b,M\in Z\Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-1;0;2;3\right\}\)

Answer 3

Câu 5:

\(ĐK:a\ne1;a\ge0\\ 1,A=\dfrac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\\ A=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\\ 2,\text{Ta có }A=1+\dfrac{2}{\sqrt{a}-1}\le1+\dfrac{2}{0-1}=1-2=-1< 0\\ \text{Mà }\left|A\right|=2\Leftrightarrow A=-2\\ \Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=-2\\ \Leftrightarrow\sqrt{a}+1=2-2\sqrt{a}\\ \Leftrightarrow\sqrt{a}=\dfrac{1}{3}\Leftrightarrow a=\dfrac{1}{9}\left(tm\right)\)