若a为根号2的小数部分,则a+1/a=?若b/a=1+根号2,且ab≠0,则(a的平方+b的平方)/2ab=?
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![若a为根号2的小数部分,则a+1/a=?若b/a=1+根号2,且ab≠0,则(a的平方+b的平方)/2ab=?](/uploads/image/z/5202658-10-8.jpg?t=%E8%8B%A5a%E4%B8%BA%E6%A0%B9%E5%8F%B72%E7%9A%84%E5%B0%8F%E6%95%B0%E9%83%A8%E5%88%86%2C%E5%88%99a%2B1%2Fa%3D%3F%E8%8B%A5b%2Fa%3D1%2B%E6%A0%B9%E5%8F%B72%2C%E4%B8%94ab%E2%89%A00%2C%E5%88%99%EF%BC%88a%E7%9A%84%E5%B9%B3%E6%96%B9%2Bb%E7%9A%84%E5%B9%B3%E6%96%B9%EF%BC%89%2F2ab%3D%3F)
若a为根号2的小数部分,则a+1/a=?若b/a=1+根号2,且ab≠0,则(a的平方+b的平方)/2ab=?
若a为根号2的小数部分,则a+1/a=?若b/a=1+根号2,且ab≠0,则(a的平方+b的平方)/2ab=?
若a为根号2的小数部分,则a+1/a=?若b/a=1+根号2,且ab≠0,则(a的平方+b的平方)/2ab=?
∵√2的整数部分为1,所以小数部分为a=√2-1
则a+1/a=√2-1+1/(√2-1)=√2-1+(√2+1)/[(√2-1)(√2+1)]=√2-1+√2+1=2√2
b/a=1+√2→b=(1+√2)a=(1+√2)×(√2-1)=1
(a²+b²)/(2ab)
=[(√2-1)²+1²]/[2×(√2-1)×1]
=(2-2√2+1+1)/(2√2-2)
=(4-2√2)/(2√2-2)
=(2-√2)/(√2-1)
=[(2-√2)(√2+1)]/[(√2-1)(√2+1)]
=(2√2+2-2-√2)/(2-1)
=√2
1)因为1<√2<2
所以a=√2-1
a+1/a
=(√2-1)+1/(√2-1)
=√2-1+√2+1
=2√2
2)由b/a=1+√2,得,
b=a(1+√2)
=(√2-1)(1+√2)
=1
所以(a^2+b^2)/2ab
=[(√2-1)^2+1]/2(√2-1)
=(4-2√2...
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1)因为1<√2<2
所以a=√2-1
a+1/a
=(√2-1)+1/(√2-1)
=√2-1+√2+1
=2√2
2)由b/a=1+√2,得,
b=a(1+√2)
=(√2-1)(1+√2)
=1
所以(a^2+b^2)/2ab
=[(√2-1)^2+1]/2(√2-1)
=(4-2√2)/2(√2-1)
=(2-√2)/(√2-1)
=(2-√2)(√2+1)
=2√2+2-2-√2
=√2
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因为根号2的整数部分为1,所以a为根号2的小数部分为(根号2)-1,然后带入分母有理化就可以算出结果。第二问你也可以直接设a=1个单位,则b=1+根号2,带入计算即可。