先化简再求值 其中A是方程2X²+X-3＝0

解（1）由韦达定理可得sinθ+cosθ=(√3+1)/2两边平方可得sin2θ=√3/2θ∈[0,2π),所以2θ=π/3或2π/3所以θ=π/6或π/3(2)由韦达定理可得sinθcosθ=m/2

a是方程x^2+3x-6的根则有a^2+3a-6=0a^2+3a=6（a^2-4a+4分之a^2-4—2-a分之1）÷a^2-2a分之2=（(a+2)(a-2)/(a-2)^2+1/(a-2)）*(a

原式={[(a+2)(a-2)/(a-2)^2+[(a-2)^2/1/(a-2)]}×a(a-2)=[(a+2)/(a-2)+1/(a-2)]×a(a-2)=[(a+3)/(a-2)]×a(a-2)=

化简：原式=a/(a+1)-(a+3)/（a-1）*（a-1)^2/[(a+2)^2-1]=a/(a+1)-[（a-1)*(a+3)]/[(a+2-1)*(a+2+1)]=a/(a+1)-(a-1)/

化简：原式=a/(a+1)-(a+3)/（a-1）*（a-1)^2/[(a+2)^2-1]=a/(a+1)-[（a-1)*(a+3)]/[(a+2-1)*(a+2+1)]=a/(a+1)-(a-1)/

原式=[m-3/3m(m-2)]÷﹛[(m+2)(m-2)-5]/m-2﹜=[m-3/3m(m-2)]÷[(m+3)(m-3)/m-2]=[m-3/3m(m-2)]×[m-2/(m+3)(m-3)]=

(a的平方-4a+4)分之(a的平方-4)-(2-a)分之1)除以(a平方-2a)分之2=[(a-2)的平方分之(a-2)(a+2)+(a-2)分之1)]除以a(a-2)分之2=2(a-2)分之(a+

[(a^2-4)/(a^2-4a+4)-a/(2-a)]÷[2/(a^2-2a)]=[(a+2)(a-2)/(a-2)^2-a/(2-a)]*[(a^2-2a)/2]=[(a+2)/(a-2)-a/(

x=ax²+2x=1原式=1/(a+1)-(a+3)/(a+1)(a-1)×(a-1)²/(a+1)(a+3)=1/(a+1)-(a-1)/(a+1)²=(a+1-a+1

a方程x平方加3x加1等于0的根．a=-3±√(3²-2)/2=-3±1/2√14(a平方减4a+4)分之a平方-4减a平方-2a分之1)除以a平方-2a分之2=[(a-2)(a+2)/(a

估计楼主的意思应该是这样,希望下次提问能知道用图片来表明这种题目

1）1+X/1-X÷(X-2X/1-X),=(1+x)/(1-x)÷(-x-x^2)/(1-x)=(1+x)/(1-x)×(1-x)/[-x(x+1)]=-1/x=-1/22)(a^2+b^2/ab+

x²+2x+1=10（x+1)²=10x+1=3或x+1=-3所以x=2或x=-4【（x²+4)/x-4】÷【（x²-4)/(x²+2x)】=【（x&

 再问：太谢谢了~

原式=[(a+2)(a-2)/(a-2)²+1/(a-2)]×a(a-2)=[(a+2)/(a-2)+1/(a-2)]×a(a-2)=a(a+2)+a=a²+3a=-1

[(a^2-4)\(a^2-4a+4)-1/(2-a)]÷2/(a^2-2a)=[(a+2)(a-2)/(a-2)^2+1/(a-2)])÷2/[a(a-2)]=(a+3)/(a-2)*[a(a-2)

x²-3x+2=0(x-1)(x-2)=0因为分母x-1≠0所以x-2=0x=2原式=x(x-1)/(x-1)*(x+1)(x-1)/(x-1)²=x*(x+1)/(x-1)=2*

你的表达式m-3/m的平方-2m/(m+2-5/m-2)不清楚,最好是把图片发上来再问：m-3/m的平方-2m除以（m+2-5/m-2）知道怎么做了嘻嘻嘻再答：还是不清楚。猜测如下：m是方程x

(5x+2-x+2)÷x-32x2+4x=（5x+2-x-21）•2x(x+2)x-3=5-(x+2)(x-2)x+2•2x(x+2)x-3=-(x+3)(x-3)x+2•2x(x+2)x-3=-2x