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初中因式分解题目:若实数x,y,z满足(x%z)&#178...

(1)原式=a(x²+2a+a²)=a(x+a)² (2)原式=3a²(2a - 3) (3)原式=16x²y²z²(1+3x² - 6z²) (4)原式=0,(5)原式=0, (6)原式=2ab(a+b)(2 - 3a) (7)原式=(x+10)(x-10) (8)原...

xy+xz-y²+z² =x(y+z)+(z+y)(z-y) =(y+z)(x+z-y)

x²-y²-z²-2yz+1-2x=x²-2x+1-(y²+2yz+z²)=(x-1)²-(y+z)² =(x-y-z-1)(x+y+z-1)

解:这是轮换对称式 当y=z时 原式=0+(z²-x²)(1+z²)(1+zx)+(x²+z²)(1+z²)(1+zx)=0 所以原式有因式(y-z),同理它应还有因式(z-x)(x-y) ∴原式有因式(y-z)(z-x)(x-y) 用这个因式除原式所得的商式应是一个三次轮换式(非...

x³+y³+z³-3xyz =x³+3x²y+3xy²+y³+z³-3x²y-3xy²-3xyz =(x+y)³+z³-3xy(x+y+z) =(x+y+z)(x²+2xy+y²-xz-yz-3xy) =(x+y+z)(x²+y²+z²-xy-yz-xz)

x³+y³+z³-3xyz =(x+y+z)(x²+y²+z²-xy-xz-yz) (这用到的是公式a³+b³+c³-3abc=(a²+b²+c²-ab-ac-bc))

9(x-y)²-y², =[3(x-y)]²-y² =[3(x-y)+y][3(x-y)-y] =(3x-2y)(3x-4y) 9a²-4(b+c)², =(3a)²-[2(b+c)]² =[3a+2(b+c)][3a-2(b+c)] =(3a+2b+2c)(3a-2b-2c) (x+y+z)²-(x-y-z)² =[(x+y+z)+(x-...

x³+y³+z³-3xyz =x³+3x²y+3xy²+y³+z³-3x²y-3xy²-3xyz =(x+y)³+z³-3xy(x+y+z) =(x+y+z)(x²+2xy+y²-xz-yz-3xy) =(x+y+z)(x²+y²+z²-xy-yz-xz)

-1/2x²y³-x²y+2x³y²z =-1/2x²y( y²+2-4xyz)

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