若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
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若三条之线2x+3y+8=0,x-y-1=0和x+ky=0相交于一点,求k值
{x+y=1 ,xy=-6{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0
已知4份之3x+2y=6份x-y+1=5份之2x+y,求x与y的值
解下列方程组:{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0
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若x-y不等于0,2x-3y=0,则分式(10x-11y)/(x-y)的值是()?(10x-11y)/(x-y) =[(8x-8y)+(2x-3y)]/(x-y) =8为什么8X-8Y=8?(10x-11y)/(x-y)=[(8x-8y)+(2x-3y)]/(x-y)=8+(2x-3y)]/(x-y)=8+0=8