Câu hỏi của nguyen minh huyen

Question

Chứng minh$a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2$$b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}$$c,2\sqrt{2}\left(3-\sqrt{2}\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9$\(d,\sqrt{\frac{4}{\left(2-\sq...

💋Bevis💋 · Accepted Answer

$a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2$Ta có:$VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}$$=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}$$=|2-\sqrt{5}|-\sqrt{5}$$=\sqrt{5}-2-\sqrt{5}$$=-2=VP\left(đpcm\right)$$b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}$Ta có:$VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}$$=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}$$=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}$$=\frac{3+2\sqrt{2}}{2-1}$$=3+2\sqrt{2}=VP\left(đpcm\right)$Câu trả lời: $a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2$Ta có:$VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}$$=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}$$=|2-\sqrt{5}|-\sqrt{5}$$=\sqrt{5}-2-\sqrt{5}$$=-2=VP\left(đpcm\right)$$b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}$Ta có:$VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}$$=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}$$=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}$$=\frac{3+2\sqrt{2}}{2-1}$$=3+2\sqrt{2}=VP\left(đpcm\right)$

💋Bevis💋 · Answer

c,Bạn xem lại đề$d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8$Ta có:$VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}$$=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}$$=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}$$=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}$$=\frac{8}{5-4}$$=8=VP\left(đpcm\right)$Câu trả lời: c,Bạn xem lại đề$d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8$Ta có:$VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}$$=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}$$=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}$$=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}$$=\frac{8}{5-4}$$=8=VP\left(đpcm\right)$

💋Bevis💋 · Answer

$e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8$$VT=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}$$=\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{\left(1-\sqrt{5}\right)^2}\right)$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)|1-\sqrt{5}|$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2$$=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)$$=\left(3+\sqrt{5}\right).2\left(3-\sqrt{5}\right)$$=[3^2-\left(\sqrt{5}\right)^2].2$$=4.2=8=VP\left(đpcm\right)$Câu trả lời: $e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8$$VT=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}$$=\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{\left(1-\sqrt{5}\right)^2}\right)$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)|1-\sqrt{5}|$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)$$=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2$$=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)$$=\left(3+\sqrt{5}\right).2\left(3-\sqrt{5}\right)$$=[3^2-\left(\sqrt{5}\right)^2].2$$=4.2=8=VP\left(đpcm\right)$

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