（2x-3y-1）（2x+3y+1） 计算题.

(2X+3y+1)^2+/4x-9y-8/=0
所以2x+3y+1=0
4x-9y-8=0
所以x=1/2 y=-2/3

由已知得
3x+y=2
2x+3y=-1
解得x=1,y=-1

(2x+3y-1)^2-(2x+3y-1)(2x+3y+1)
=(2x+3y-1)[(2x+3y-1)-(2x+3y+1)]
=-2(2x+3y-1)

您好!
由题得｜3x+y-2|=0,(2x+3y+1)^2=0
故3x+y-2=0
2x+3y+1=0
解得x=1
y=-1
谢谢~~

1.
38.5^2-36.5^2
=(38.5+36.5)(38.5-36.5)
=75*2
=150
2.
(2x-3y-1)(2x+3y+1)
=[2x-(3y+1)][2x+(3y+1)]
=4x^2-(3y+1)^2
=4x^2-(9y^2+6y+1)
=4x^2-9y^2-6y-1
3.
(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(1/2)(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(1/2)(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(1/2)(3^4-1)(3^4+1)(3^8+1)(3^16+1)
=(1/2)(3^8-1)(3^8+1)(3^16+1)
=(1/2)(3^16-1)(3^16+1)
=(1/2)(3^32-1)
=3^32/2-1/2

|..| 绝对值≥0
(..)^2 平方值≥0
所以
{3x+y—2=0 ①
{2x+3y+1=0 ②
①*3 - ②得
9x+3y-6-2x-3y-1=0
7x=7
x=1
所以y=-1

显然‍3x+y-2的绝对值≥0,（2x+3y+1）的平方≥0
满足3x+y-2的绝对值+（2x+3y+1）的平方＝0
所以3x+y-2=0,2x+3y+1=0
联立解得:x=1,y=-1

（1）(5a+6b-7c)^2-(5a-6b+7c)^2=【(5a+6b-7c)+(5a-6b+7c)】×【(5a+6b-7c)-(5a-6b+7c)】
=10a（12b-14c）=120ab-140ac
（2)（2x-3y-1)(2x+3y+1)=【2x-（3y+1)】×【2x+(3y+1)】=4x^2-(3y+1)^2=4x^2-9y^2-6y-1
(3)没看懂,另请高人吧!
(4)已知x^m=9,x^n=6.x^k=4求x^m-2n-2k
解x^(m-2n+2k)=x^m/(x^n)^2*(x^k)^2=9/36*16=4
（5）a^a-3=1 a^a-3×a^3=1×a^3 所以a^a=a^3 所以a=3
(6)你不小心打错数了吧,7x应为7x^2：若3x^2-x=1,求6x^3+7x^2-5x+2009
因为3x^2-x=1,所以3x^2=x+1
所以6x^3+7x^2-5x+2009
=x（6x^2+7x-5）+2009
=x(2x-1）(3x+5)+2009
=（2x-1）（3x^2+5x)+2009
=（2x-1）（1+x+5x)+2009
=(2x-1)(6x+1)+2009
=12x^2-4x-1+2009
=4(3x^2-x)+2008
=4*1+2008
=2012