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9x²-y²+4y-4用因式分解怎么做

9x²-y²+4y-4=9x²-(y²-4y+4)=(3x)²-(y-2)²=(3x+y-2)(3x-y+2)

9x²-y²-4y-4=9x²-(y²+4y+4)=(3x)²-(y+2)²=平方差公式,自己算

16(x-y)²-25(x+y)² =[4(x+y)]²-[5(x-y)]² =(4x+4y+5x-5y)(4x+4y-5x+5y) =(9x-y)(-x+9y) 25(a-2b)³+4(2b-a) =25(a-2b)³-4(a-2b) =(a-2b)[25(a-2b)²-4] =(a-2b)[5(a-2b)+2][5(a-2b)-2] =(a-2b)(5a-10b...

9x²-y²-4y-4 =9x²-(y+2)² =(3x+y+2)(3x-y-2)

9x” - y” - 4y - 4 = (3x)” - [ y” + 4y + 4 ] = (3x)” - ( y + 2 )” = [ 3x - ( y + 2 ) ][ 3x + ( y + 2 ) ] = ( 3x - y - 2 )( 3x + y + 2 )

这道题用到点差法, 就是设那两点是x1 y1 x2 y2 将其带入圆的方程 然后相减 运用平方差公式 可以看出一些结果 把斜率代进去

(3x-2y)²(3x+2y)²(9x²+4y²)=[(3x+2y)(3x-2y)]^2(9x²+4y²)=(9x^2-4y^2)(9x²+4y²)=81x^4-16y^4

x8-98x4y4+y8=x8+2x4y4+y8-100x4y4 =(x4+y4)²-(10x²y²)² =(x4+y4-10x²y²)(x4+y4+10x²y²)

若√(x-y)+y²-4y+4=0,求xy的值 ∵√(x-y)+y²-4y+4=0 ∴√(x-y)+(y-2)²=0 ∴x-y=0,y-2=0 即x=y=2 ∴xy=2×2=4

(1)解原式=(x+4)(x-1) (2)解原式=(x+2y)(x-2y)+(x+2y) =(x+2y)(x-2y+1) (3)解原式=(a+b)(x+y-3x+3y) =(a+b)(4y-2x) =2(a+b)(2y-x)

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