这几道数学题怎么做?都是分式哦.(a+b)/ab-(b+c)/bc3x/(x-3)²-x/3-x1/x-1 + 1/1+xc/ab - a/bca/a²-1 + 3a+1/a²-1 + 2a+3/1-a²

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 13:27:21

这几道数学题怎么做?都是分式哦.
(a+b)/ab-(b+c)/bc
3x/(x-3)²-x/3-x
1/x-1 + 1/1+x
c/ab - a/bc
a/a²-1 + 3a+1/a²-1 + 2a+3/1-a²

(a+b)/ab-(b+c)/bc
=[c(a+b)-a(b+c)]/(abc)
=(ca+bc-ab-ac)/(abc)
=b(c-a)/(abc)
=(c-a)/ac
=1/a-1/c
3x/(x-3)²-x/(3-x)
=[3x+x(x-3)]/(x-3)²
=x²/(x-3)²
1/(x-1)+ 1/(1+x)
=[(x+1)+(x-1)]/[(x-1)(1+x)]
=2x/(x²-1)
c/ab - a/bc
=(c²-a²)/(abc)
a/(a²-1) + (3a+1)/(a²-1) + (2a+3)/(1-a²)
=[a+(3a+1)-(2a+3)]/(a²-1)
=([2a-2)/(a²-1)
=2(a-1)/[(a+1)(a-1)]
=2/(a+1)

由已知变形得:
(a+b)/ab=1/2
(b+c)/bc=1/3
(c+a)/ca=1/4
进一步变形为:
1/a+1/b=1/2········①
1/b+1/c=1/3········②
1/c+1/a=1/4········③
①+②+③,得:
2(1/a+1/b+1/c)=1/2+1/3+1/4=1...

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由已知变形得:
(a+b)/ab=1/2
(b+c)/bc=1/3
(c+a)/ca=1/4
进一步变形为:
1/a+1/b=1/2········①
1/b+1/c=1/3········②
1/c+1/a=1/4········③
①+②+③,得:
2(1/a+1/b+1/c)=1/2+1/3+1/4=13/12
1/a+1/b+1/c=13/24
上式分别减去①、②、③,得:
1/c=1/24
得:c=24,
1/a=5/24
得:a=24/5,
1/b=7/24
得:b=24/7;
所以
a+b+c
=24/5+24/7+24
=1128/35 5x/(x-3)+(3x-1)/(3-x)
=5x/(x-3)-(3x-1)/(x-3)
=(5x-3x+1)/(x-3)
=(2x+1)/(x-3)
=(Ax+B)/(x-3),
A=2
B=1 1.f(x-1/x)=1/(1/x²+x²)=1/((x-1/x)²+2)
f(x)=1/(x²+2)
2.f(g(x))=a^(g²(x))
a^(g²(x))=1-x
g²(x)=loga(1-x)
g(x)=根号下loga(1-x)
3.f(x²-1)=lg(x²-1+1)/(x²-1-1)
f(x)=lg(x+1)/(x-1)
f(φ(x))=lg(φ(x)+1)/(φ(x)-1)
(φ(x)+1)/(φ(x)-1)=x
φ(x)+1=φ(x)x-x
φ(x)(x-1)=x+1
φ(x)=(x+1)/(x-1)
4.f(x)-1=x/(1-x)-1=(2x-1)/(1-x),f(x)/(f(x)-1)=x/(2x-1)
f(f(x)/(f(x)-1))=(x/(2x-1))/(1-x/(2x-1))
=(x/(2x-1))/((x-1)/(2x-1))
=x/(x-1) 这个不等式不成立的。你可以把a=b=c=√3/3代入试一下:
左边=3^(5/4),而右边=3^(7/4)
很明显:左边<右边 不对的。把右边的3换成√3就对了。即:
√(a/bc)+√(b/ac)+√(c/ab)≥√3(√a+√b+√c)
证明:
左边通分,原不等式等价于:
(a+b+c)/√(abc)≥√3(√a+√b+√c)
两边平方,不等式等价于:
(a+b+c)^2/(abc)≥3(√a+√b+√c)^2
左边乘以(ab+bc+ca),不等式等价于:
(ab+bc+ca)(a+b+c)^2/(abc)≥3(√a+√b+√c)^2
即:(1/a+1/b+1/c)(a+b+c)^2≥3(√a+√b+√c)^2……①
根据平均值不等式:
1/a+1/b+1/c≥3/(abc)^(1/3)……②
a+b+c≥3(abc)^(1/3)……③
根据柯西不等式:
3(a+b+c)
=(1^2+1^2+1^2)[(√a)^2+(√b)^2+(√c)^2]
≥(√a+√b+√c)^2
即:a+b+c≥[(√a+√b+√c)^2]/3……④ a²-3a+1=0
a-3+ 1/a=0
a+1/a=3
a²+1/a²=(a+1/a)^2-2=7
②③④三式相乘即得①式。
故√(a/bc)+√(b/ac)+√(c/ab)≥√3(√a+√b+√c)得证。
等号成立的条件是a=b=c=√3/3
=-x/(1-x)
=-f(x)

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(c-a)/ac x的平方除以X-3括号的平方

都是化简啊,简单
找公因式即可!
1.公因式abc,化简得(c-a)/ac
2.公因式(x-3)²
3.公因式x^2-1
4.公因式abc
5,公因式a²-1