Câu hỏi của manh

Question

a : $\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$b : $\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}$c : $\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}$

HT.Phong (9A5) · Accepted Answer

a) $\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|$$=\sqrt{2}-1+2-\sqrt{2}$$=1$b) $\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|$$=2\sqrt{6}-3-5+2\sqrt{6}$$=4\sqrt{6}-8$c) $\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}$$=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}$$=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}$$=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|$$=\sqrt{6}-1+3-\sqrt{6}$$=2$Câu trả lời: a) $\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|$$=\sqrt{2}-1+2-\sqrt{2}$$=1$b) $\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|$$=2\sqrt{6}-3-5+2\sqrt{6}$$=4\sqrt{6}-8$c) $\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}$$=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}$$=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}$$=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|$$=\sqrt{6}-1+3-\sqrt{6}$$=2$

Toru · Answer

$a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|$$=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}$$=\left|\sqrt{2}-1\right|+2-\sqrt{2}$$=\sqrt{2}-1+2-\sqrt{2}$$=1$$---$$b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|$$=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}$$=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}$$=2\sqrt{6}-3-5+2\sqrt{6}$$=4\sqrt{6}-8$$---$$c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}$$=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}$$=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}$$=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|$$=\sqrt{6}-1+3-\sqrt{6}$$=2$#$Toru$Câu trả lời: $a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|$$=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}$$=\left|\sqrt{2}-1\right|+2-\sqrt{2}$$=\sqrt{2}-1+2-\sqrt{2}$$=1$$---$$b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}$$=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|$$=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}$$=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}$$=2\sqrt{6}-3-5+2\sqrt{6}$$=4\sqrt{6}-8$$---$$c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}$$=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}$$=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}$$=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|$$=\sqrt{6}-1+3-\sqrt{6}$$=2$#$Toru$

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