(3.14-兀)0次方乘以(二分之一)负一次方--根号3-1分之2+(根号三-2)2009次方乘以(根号三+2)2010次方

2024-12-05 07:07:18
推荐回答(5个)
回答1:

解;(3.14-兀)0次方=1

    (二分之一)负一次方=2

       根号3-1分之2=2(根号3+1)÷(3-1)=根号3+1

     (根号三-2)2009次方乘以(根号三+2)2010次方=[(根号三-2)×(根号三+2)] 2009次方 

                                                   × (根号三+2)=(3-4)2009次方 ×(根号三+2)=-(根号3+2)

        所以

     (3.14-兀)0次方乘以(二分之一)负一次方-根号3-1分之2+(根号三-2)2009次方乘以

     (根号三+2)2010次方=1×2-(根号3+1)-(根号3+2)=-2倍根号3-1

答;______________ 

【点拨1】:{根号3-1分之2}  与   {根号三-2)2009次方乘以(根号三+2)2010次方}  不是一个

                    整体 因为他们之间还有“+”呢,在小学的四项运算法则中就知道先乘除再加减,所以

                   这种说法不对

【点拨2】:这里看作整体是一个化整为零的做法,利用变换组合的逻辑思维操作的,你可以看

                   一下插图就明白了!希望能帮助到你!

回答2:

(3.14-π)º· (1/2)^(-1)-[-2/(√3-1)]+(√3-2)^2009·(√3+2)^2010=2-(√3+1)+(3-4)^2009·(√3+2)=2-√3-1-(√3+2)=-2√3-1

回答3:

(3.14-兀)0次方=1;
(二分之一)负一次方=2;
其他的没看明白

回答4:

(3.14-π)^0*(1/2)^(-1)-2/(√3-1)+(√3-2)^2009*(√3+2)^2010
= 1*(1/2)^(-1)-2/(√3-1)+(√3-2)^2009*(√3+2)^2010
= 2^1-2/(√3-1)+(√3-2)^2009*(√3+2)^2010
= 2-2/(√3-1)+(√3-2)^2009*(√3+2)^2010
= 2-2(√3+1)/(√3-1)(√3+1)+(√3-2)^2009*(√3+2)^2010
= 2-2(√3+1)/2+(√3-2)^2009*(√3+2)^2010
= 2-(√3+1)+(√3-2)^2009*(√3+2)^2010
= 2-(√3+1)+(√3-2)^2009*(√3+2)^2009*(√3+2)
= 2-(√3+1)+[(√3-2)(√3+2)]^2009*(√3+2)
= 2-√3-1+(-1)^2009*(√3+2)
= 1-√3-(√3+2)
= 1-√3-√3-2
=-2√3-1

回答5:

写的太乱了

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