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解:a²-3a+1=0
则 a²=3a-1 a很显然不等于0
所以 要想 求a³/a^6+1
我们不妨先求其倒数
即 (a^6+1)/a³
=a³+1/a³ (立方和公式)
=(a+1/a)(a²-1+1/a²) (通分)
=(a²+1)/a x (a²-1+1/a²) (将a²=3a-1 带入)
=3a/a x [(3a-1)-1+1/(3a-1)] (后面的通分)
=3 x [(3a-1)²+1-(3a-1)]/(3a-1) (将a²=3a-1 带入并化简)
=3x(18a-6)/(3a-1)
=3x6
=18
所以 a³/(a^6+1)=1/ (a^6+1)/a³
=1/18
则 a²=3a-1 a很显然不等于0
所以 要想 求a³/a^6+1
我们不妨先求其倒数
即 (a^6+1)/a³
=a³+1/a³ (立方和公式)
=(a+1/a)(a²-1+1/a²) (通分)
=(a²+1)/a x (a²-1+1/a²) (将a²=3a-1 带入)
=3a/a x [(3a-1)-1+1/(3a-1)] (后面的通分)
=3 x [(3a-1)²+1-(3a-1)]/(3a-1) (将a²=3a-1 带入并化简)
=3x(18a-6)/(3a-1)
=3x6
=18
所以 a³/(a^6+1)=1/ (a^6+1)/a³
=1/18
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a²-3a+1=0
∴a²=3a-1
a³=a(3a-1)=3a²-a=3(3a-1)-a=8a-3
a³/(a的六次方+1)
=(8a-3)/[(8a-3)²+1]
=(8a-3)/[64a²-48a+10]
=(8a-3)/(192a-48a-54)
=(8a-3)/(144a-54)
=(8a-3)/18(8a-3)
=1/18
∴a²=3a-1
a³=a(3a-1)=3a²-a=3(3a-1)-a=8a-3
a³/(a的六次方+1)
=(8a-3)/[(8a-3)²+1]
=(8a-3)/[64a²-48a+10]
=(8a-3)/(192a-48a-54)
=(8a-3)/(144a-54)
=(8a-3)/18(8a-3)
=1/18
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解:b²-4ac=(-3)²-4×1×1=5
则√(b²-4ac)=√5
则-b+√(b²-4ac)=3+√5,-b-√(b²-4ac)=3-√5
∴a1=(3+√5)/2,a2=(3-√5)/2
当a=(3+√5)/2时,
原式=[(3+√5)/2]³÷{[(3+√5)/2]^6+1}=(9+4√5)÷[(162+72√5)/16]=(72+32√5)/(81+36√5)
当a=(3-√5)/2时
原式=[(3-√5)/2]³÷{[(3-√5)/2]^6+1}=(9-4√5)÷[(162-72√5)/16]=(72-32√5)/(81+36√5
则√(b²-4ac)=√5
则-b+√(b²-4ac)=3+√5,-b-√(b²-4ac)=3-√5
∴a1=(3+√5)/2,a2=(3-√5)/2
当a=(3+√5)/2时,
原式=[(3+√5)/2]³÷{[(3+√5)/2]^6+1}=(9+4√5)÷[(162+72√5)/16]=(72+32√5)/(81+36√5)
当a=(3-√5)/2时
原式=[(3-√5)/2]³÷{[(3-√5)/2]^6+1}=(9-4√5)÷[(162-72√5)/16]=(72-32√5)/(81+36√5
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a²-3a+1=0,
A-3+1/A=0 ==>A+1/A=3 ==>(A+1/A)^2=3 ==>A^2+2+1/A^2=9 ==>A^2+1/A^2=7
A^3/(A^6+1)=1/(A^3+1/A^3)=1/((A+1/A)(A^2-1+1/A^2)=1/(3*(7-1)=1/18
A-3+1/A=0 ==>A+1/A=3 ==>(A+1/A)^2=3 ==>A^2+2+1/A^2=9 ==>A^2+1/A^2=7
A^3/(A^6+1)=1/(A^3+1/A^3)=1/((A+1/A)(A^2-1+1/A^2)=1/(3*(7-1)=1/18
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a^2=3a-1
a^6+1=(3a-1)^3+1
=(27a^3-1-27a^2+9a)+1
=27a^3-27a^2+9a
=9a(3a^2-3a+1)
=9a[3a^2-(3a-1)]
=9a*2a^2
a^3/(a^6+1)=a3/(9a*2a^2)=1/18
a^6+1=(3a-1)^3+1
=(27a^3-1-27a^2+9a)+1
=27a^3-27a^2+9a
=9a(3a^2-3a+1)
=9a[3a^2-(3a-1)]
=9a*2a^2
a^3/(a^6+1)=a3/(9a*2a^2)=1/18
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