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Question

Cứu mình phần b

乇尺尺のﾚ · Accepted Answer

$a)C=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{2}{x-1}$$b)x=\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}-1}-\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}+1}\
=\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}+1\right)}{1+\sqrt{7}-1}-\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}-1\right)}{1+\sqrt{7}-1}\
=\sqrt{1+\sqrt{7}}+1-\sqrt{1+\sqrt{7}}+1\
=2$Thay $x=2$  vào $C$$\dfrac{2}{2-1}=\dfrac{2}{1}=2$Câu trả lời: $a)C=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\
=\dfrac{2}{x-1}$$b)x=\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}-1}-\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}+1}\
=\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}+1\right)}{1+\sqrt{7}-1}-\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}-1\right)}{1+\sqrt{7}-1}\
=\sqrt{1+\sqrt{7}}+1-\sqrt{1+\sqrt{7}}+1\
=2$Thay $x=2$  vào $C$$\dfrac{2}{2-1}=\dfrac{2}{1}=2$

乇尺尺のﾚ · Answer

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