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Đáp án+Giải thích các bước giải:
`A=(1+sqrt5)/(sqrt2+sqrt(3+sqrt5)) +(1-sqrt5)/(sqrt2-sqrt(3-sqrt5))`
`=(sqrt2+sqrt10)/(2+sqrt(6+2sqrt5)) +(sqrt2-sqrt10)/(2-sqrt(6-2sqrt5))`
`=(sqrt2(1+sqrt5))/(2+sqrt(5+2sqrt5+1)) +(sqrt2(1-sqrt5))/(2-sqrt(5-2sqrt5+1))`
`=(sqrt2(1+sqrt5))/(2+sqrt((sqrt5+1)^2)) +(sqrt2(1-sqrt5))/(2-sqrt((sqrt5-1)^2))`
`=(sqrt2(1+sqrt5))/(2+sqrt5+1) +(sqrt2(1-sqrt5))/(2-sqrt5+1)`
`=(sqrt2(1+sqrt5))/(3+sqrt5) +(sqrt2(1-sqrt5))/(3-sqrt5)`
`=(sqrt2(1+sqrt5).(3-sqrt5))/(9-(sqrt5)^2) +(sqrt2(1-sqrt5).(3+sqrt5))/(9-(sqrt5)^2) `
`=(sqrt2.(3-sqrt5+3sqrt5-5)+sqrt2.(3+sqrt5-3sqrt5-5))/4`
`=(sqrt2.(2sqrt5-2-2sqrt5-2))/4`
`=((-4.)sqrt2)/4`
`=-sqrt2`
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Đáp án:
`A=-\sqrt{2}`
Giải thích các bước giải:
`a)`
`A=\frac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{\sqrt{2}.(\sqrt{2}+\sqrt{3+\sqrt{5}})}+\frac{\sqrt{2}.(1-\sqrt{5})}{\sqrt{2}.(\sqrt{2}-\sqrt{3-\sqrt{5}})}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{2}.\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{2}.\sqrt{3-\sqrt{5}}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{(\sqrt{2})^{2}.(3+\sqrt{5})}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{(\sqrt{2})^{2}.(3-\sqrt{5})}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{2.(3+\sqrt{5})}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{2.(3-\sqrt{5})}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{6-2\sqrt{5}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{5+2\sqrt{5}+1}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{5-2\sqrt{5}+1}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{(\sqrt{5})^{2}+2.\sqrt{5}.1+1^{2}}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{(\sqrt{5})^{2}-2.\sqrt{5}.1+1^{2}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{(\sqrt{5}+1)^{2}}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{(\sqrt{5}-1)^{2}}}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+|\sqrt{5}+1|}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-|\sqrt{5}-1|}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{2+\sqrt{5}+1}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-(\sqrt{5}-1)}` (vì `\sqrt{5}+1>0;\sqrt{5}-1>0`)
`=\frac{\sqrt{2}.(1+\sqrt{5})}{3+\sqrt{5}}+\frac{\sqrt{2}.(1-\sqrt{5})}{2-\sqrt{5}+1}`
`=\frac{\sqrt{2}.(1+\sqrt{5})}{3+\sqrt{5}}+\frac{\sqrt{2}.(1-\sqrt{5})}{3-\sqrt{5}}`
`=\frac{\sqrt{2}.(1+\sqrt{5}).(3-\sqrt{5})}{(3+\sqrt{5}).(3-\sqrt{5})}+\frac{\sqrt{2}.(1-\sqrt{5}).(3+\sqrt{5})}{(3+\sqrt{5}).(3-\sqrt{5})}`
`=\frac{\sqrt{2}.(3-\sqrt{5}+3\sqrt{5}-5)+\sqrt{2}.(3+\sqrt{5}-3\sqrt{5}-5)}{3^{2}-(\sqrt{5})^{2}}`
`=\frac{\sqrt{2}.(2\sqrt{5}-2-2-2\sqrt{5})}{9-5}`
`=\frac{\sqrt{2}.(-4)}{4}`
`=-\sqrt{2}`
Vậy `A=-\sqrt{2}`
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