若根号3减A分之根号2A加B
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设根号a=A,根号b=B则A^2+B^2=6,A^2B^2=4,AB=2所有为:(A-B)/(A+B)(A-B)^2=A^2-2AB+B^2=6-4=2(A+B)^2=A^2+2AB+B^2=6+4=
a+b=-3,ab=2(a+b)²=9a²+b²+2ab=9a²+b²+2*2=9a²+b²=5(a²+b²)
a=根号3减根号2分之根号3加根号2=(√3+√2)²/[(√3-√2)(√3+√2)]=﹙3+2+2√6)/(3-2)=5+2√6b=根号3加根号2分之根号3减根号2=(√3-√2)
因(根号12)+1/(根号3)+根号16=a+b*根号32(根号3)+(根号3)/3+4=a+b(根号3)将上式去分母,并整理得:7根号3+12=3a+3b*根号3---(1)(一)因a,b均为有理数
3根号a加4根号b减根号4分之a加根号b=3√a-1/2√a+4√b+√b=5/2√a+5√b如追加其它问题,
根号a加根号b分之根号a减根号b=±√[√a-√b)²/(√a+√b)²]=±√[a+b-2√ab)/(a+b+2√ab)]=±√(5-4)/(5+4)=±1/3希望你能看懂,望采
5-a*√3=2b+2/3*√3-a(a+2/3)√3=5+a-2b因为是有理数所以a+2/3=05+a-2b=0即a=-2/3b=13/6.
∵5-√3a=2b+2√3/3-a∴5-√3*a=(2b-a)+2/3*√3∵a,b是有理数,√3是无理数∴{5=2b-a{-a=2/3解得a=-2/3,b=13/6
根号3+根号2分之1-根号3-1分之2.=(根号3+根号2)(根号3-根号2)分之(根号3-根号2)-(根号3-1)(根号3+1)分之2(根号3+1)=(3-1)分之(根号3-根号2)-(3-1)分之
(√a+√b)(√a-√b)=a-b则√a-√b=(a-b)/(√a+√b)=6/(2√3)=√3选B
a加b等于负4,ab等于2a+b=-4ab=2所以a,b都小于0根号b分之a加根号a分之b=根号(ab)/(-a)+根号(ab)/(-b)=-根号(ab)(1/a+1/b)=-根号2(a+b)/ab=
a=根号二除以4+3+3根号2=11/4根号2+3b=11/4+3/2*根号2
(√a-√b)^2/(√a+√b)^2=(a+b-2√ab)/(a+b+2√ab)=(6√ab-2√ab)/(6√ab+2√ab)=4√ab/8√ab=1/2,所以(√a-√b)/(√a+√b)=√2
我把我理解得做一下首先是复述题目:(先汗一个……)a/(a-√(ab))-√b/(√a+√b)先把前面的分子式化简√a/(√a-√b)-√b/(√a+√b)通分(√a*(√a+√b)-√b*(√a-√
a/(a-√ab)-√b/(√a+√b)=(√a)²/[√a(√a-√b)]-√b/(√a+√b)=√a/(√a-√b)-√b/(√a+√b)=√a(√a+√b)/[(√a-√b)(√a+√
因为√a+√b=√5+√3①,√(ab)=√15-√3②将①两边平方,得a+b+2√(ab)=8+2√15③将②代入③,得a+b+2(√15-√3)=8+2√15所以a+b=8+2√3
化简根号a加根号b分之根号a减根号b,得(-根号ab)/(a-b)=(a+b)/(a-b)
2根号a分之b把b放到根号里b根号b分之a=根号a