若正数a,b满足a2/(a4+a2+1)=1/24, b3/(b6+b3+1)=1/19,则ab/(a2+a+1)(b2
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/15 21:21:27
若正数a,b满足a2/(a4+a2+1)=1/24, b3/(b6+b3+1)=1/19,则ab/(a2+a+1)(b2+b+1)=
![若正数a,b满足a2/(a4+a2+1)=1/24, b3/(b6+b3+1)=1/19,则ab/(a2+a+1)(b2](/uploads/image/z/18029556-36-6.jpg?t=%E8%8B%A5%E6%AD%A3%E6%95%B0a%2Cb%E6%BB%A1%E8%B6%B3a2%2F%28a4%2Ba2%2B1%29%3D1%2F24%2C+b3%2F%28b6%2Bb3%2B1%29%3D1%2F19%2C%E5%88%99ab%2F%28a2%2Ba%2B1%29%28b2)
a² /(a⁴+a²+1)=1/24
(a⁴+a²+1)/a²=24
a²+1+1/a²=24
(a+1/a)²-1=24
(a+1/a)²=25
a是正数
a+1/a=5
b^3/(b^6+b^3+1)=1/19
(b^6+b^3+1)/b^3=19
b^3+1+1/b^3=19
(b+1/b)^3-3b-3/b+1=19
(b+1/b)^3-3(b+1/b)-18=0
(b+1/b)^3-3(b+1/b)^2+3(b+1/b)^2-3(b+1/b)-18=0
(b+1/b)^2(b+1/b-3)+3[(b+1/b)^-(b+1/b)-6]=0
(b+1/b)^2(b+1/b-3)+3(b+1/b-3)(b+1/b+2)=0
(b+1/b-3)[(b+1/b)^2+3(b+1/b+2)]=0
(b+1/b)^2+3(b+1/b+2)
=(b+1/b)^2+3(b+1/b)+6
=(b+1/b+3/2)^2-9/4+6>0
所以b+1/b-3=0
b+1/b=3
(a²+a+1)(b²+b+1)/ab
=(a+1+1/a)(b+1+1/b)
=(5+1)(3+1)
=24
再问: 题为ab/(a2+a+1)(b2+b+1),所以答案应为1/24.
(a⁴+a²+1)/a²=24
a²+1+1/a²=24
(a+1/a)²-1=24
(a+1/a)²=25
a是正数
a+1/a=5
b^3/(b^6+b^3+1)=1/19
(b^6+b^3+1)/b^3=19
b^3+1+1/b^3=19
(b+1/b)^3-3b-3/b+1=19
(b+1/b)^3-3(b+1/b)-18=0
(b+1/b)^3-3(b+1/b)^2+3(b+1/b)^2-3(b+1/b)-18=0
(b+1/b)^2(b+1/b-3)+3[(b+1/b)^-(b+1/b)-6]=0
(b+1/b)^2(b+1/b-3)+3(b+1/b-3)(b+1/b+2)=0
(b+1/b-3)[(b+1/b)^2+3(b+1/b+2)]=0
(b+1/b)^2+3(b+1/b+2)
=(b+1/b)^2+3(b+1/b)+6
=(b+1/b+3/2)^2-9/4+6>0
所以b+1/b-3=0
b+1/b=3
(a²+a+1)(b²+b+1)/ab
=(a+1+1/a)(b+1+1/b)
=(5+1)(3+1)
=24
再问: 题为ab/(a2+a+1)(b2+b+1),所以答案应为1/24.
[1/(a-b)-(a+b)/(a2+ab+b2)+ab/(b3-a3)]×(a3-b3)
是不是这样的a6-b6=(a3+b3)(a3-b3)=(a+b)((a2-ab+b2)(a-b)(a2+ab+b2)=(
横栏A,B竖栏1,2,3,4,5,6,7等等想算出A1/B1+A2/B2+A3/B3+A4/B4+A5/B5+A6/B6
已知a-b=1,a2+b2=13,求(a3-2b3)-(a2b-2ab2)-(ab2-b3)的值
已知ab≠0,求证:a+b=1的充要条件是a3+b3+ab-a2-b2=0.
已知a+b+c=1求证 a3+b3+c3>=1/3(a2+b2+c2)
已知a+b=1,a2+b2=2,求a3+b3的值
证明:a3+b3=(a+b)(a2-ab+b2)
:(a-b)(a2 ab b2)=a3-b3 怎么解
a3+b3=(a+b)(a2-ab+b2) ?
当a=1/3,b=0.2时,求a3-b3/a2+ab+b2的值
已知实数a,b分别满足a3-3a2+5a=1,b3-3b2+5b=5,则a+b的值为______.