Question

高一对数题

根号（lg3）^2-lg9+1 * （lg根号27+lg8-lg根号1000）/（lg0.3）(lg1.2)

根号（lg3）^2-lg9+1，整一个式子都在根号下
写全过程，谢谢

Answer 1

原式=√(lg²3-2lg3+1)*(lg3√3+3lg2-1/2*lg1000)/(lg0.3)(lg1.2)
=√(lg3-1)²(lg3√3+3lg2-3/2)/(lg0.3)(lg4*3/10)
=(1-lg3)(lg3√3+3lg2-3/2/(lg3-1)(lg4+lg3-1)
=-(3/2)(2/3*lg3√3+2lg2-1/(lg4+lg3-1)
=-(3/2)(lg3+lg4-1/(lg4+lg3-1)
=-3/2

Answer 2

x=2/(5-√7)
=2(5+√7)/(5-√7)
=2(5+√7)/(25-7)
所以x=(5+√7)/9
1、化简.
51a^n+1·b^n+2·c^n+3/68a^n-1·b^n-2·c^n-3
已经是最简了

2、若x^2n=2,求(2x^3n)²-(3x^n)²的值
=8x^6n-9x^2n
=8(x^2n)³-9x^2n
=8*8-9*2
=46

3、分解因式：a²(x²-y²)+b²(y²-x²)
=a²(x²-y²)-b²(x²-y²)
=(a²-b²)(x²-y²)
=(a+b)(a-b)(x+y)(x-y)