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HT.Phong (9A5) · Accepted Answer

1) $\sqrt{11-6\sqrt{2}}-\sqrt{6-4\sqrt{2}}$$=\sqrt{9-6\sqrt{2}+2}-\sqrt{4-4\sqrt{2}+2}$$=\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{2}\cdot\left(\sqrt{2}\right)^2}$$=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}$$=\left|3-\sqrt{2}\right|-\left|2-\sqrt{2}\right|$$=3-\sqrt{2}-2+\sqrt{2}=1$2) $\sqrt{\left(1-\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}$$=\left|1-\sqrt{2}\right|-\left|2-\sqrt{2}\right|$$=\sqrt{2}-1+2-\sqrt{2}$$=1$3) $\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}$$=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{9+12\sqrt{2}+8}$$=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}$$=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}$$=\left|2\sqrt{2}-1\right|-\left|3+2\sqrt{2}\right|$$=2\sqrt{2}-1-3-2\sqrt{2}=-4$4) $\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}$$=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{9-6\sqrt{5}+5}$$=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}$$=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}$$=\left|2-\sqrt{5}\right|+\left|3-\sqrt{5}\right|$$=\sqrt{5}-2+3-\sqrt{5}=1$5) $\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}$$=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{18+6\sqrt{2}+1}$$=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{2}\right)^2+2\cdot3\sqrt{2}\cdot1+1^2}$$=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{2}+1\right)^2}$$=\left|4-3\sqrt{2}\right|-\left|3\sqrt{2}+1\right|$$=3\sqrt{2}-4-3\sqrt{2}-1=-5$6) $\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{2-2\sqrt{2}+1}+\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}$$=\sqrt{2}-1+2-\sqrt{2}=1$Câu trả lời: 1) $\sqrt{11-6\sqrt{2}}-\sqrt{6-4\sqrt{2}}$$=\sqrt{9-6\sqrt{2}+2}-\sqrt{4-4\sqrt{2}+2}$$=\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{2}\cdot\left(\sqrt{2}\right)^2}$$=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}$$=\left|3-\sqrt{2}\right|-\left|2-\sqrt{2}\right|$$=3-\sqrt{2}-2+\sqrt{2}=1$2) $\sqrt{\left(1-\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}$$=\left|1-\sqrt{2}\right|-\left|2-\sqrt{2}\right|$$=\sqrt{2}-1+2-\sqrt{2}$$=1$3) $\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}$$=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{9+12\sqrt{2}+8}$$=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}$$=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}$$=\left|2\sqrt{2}-1\right|-\left|3+2\sqrt{2}\right|$$=2\sqrt{2}-1-3-2\sqrt{2}=-4$4) $\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}$$=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{9-6\sqrt{5}+5}$$=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}$$=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}$$=\left|2-\sqrt{5}\right|+\left|3-\sqrt{5}\right|$$=\sqrt{5}-2+3-\sqrt{5}=1$5) $\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}$$=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{18+6\sqrt{2}+1}$$=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{2}\right)^2+2\cdot3\sqrt{2}\cdot1+1^2}$$=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{2}+1\right)^2}$$=\left|4-3\sqrt{2}\right|-\left|3\sqrt{2}+1\right|$$=3\sqrt{2}-4-3\sqrt{2}-1=-5$6) $\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{2-2\sqrt{2}+1}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}$$=\sqrt{2}-1+2-\sqrt{2}=1$

Nguyễn Lê Phước Thịnh · Answer

1: =3-căn 2-2+căn 2=12: $=\sqrt{2}-1+2-\sqrt{2}=1$3: $=2\sqrt{2}-1-3-2\sqrt{2}=-4$4: $=\sqrt{5}-2+3-\sqrt{5}=1$5: $=3\sqrt{2}-4-3\sqrt{2}-1=-5$6: $=\sqrt{2}-1+2-\sqrt{2}=1$Câu trả lời: 1: =3-căn 2-2+căn 2=12: $=\sqrt{2}-1+2-\sqrt{2}=1$3: $=2\sqrt{2}-1-3-2\sqrt{2}=-4$4: $=\sqrt{5}-2+3-\sqrt{5}=1$5: $=3\sqrt{2}-4-3\sqrt{2}-1=-5$6: $=\sqrt{2}-1+2-\sqrt{2}=1$

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