X的3次方减2乘以X的平方加4等于0这个方程怎样解-搜答案-神马作文网

你好!13x²-6xy+y²-4x+1=0(9x²-6xy+y²)+(4x²-4x+1)=0(3x-y)²+(2x-1)²=0∴3

4的x次方减5乘以2的x次方加4等于0（2的x次方）²-5×（2的x次方）+4=0（2的x次方-1)(2的x次方-4=02的x次方=1；2的x次方=4所以：x=0；x=2

∵3X的n加1方乘以y的4次方与2分之一X的3次方乘以y的m减2方是同类项,∴n+1=34=m-2m=2n=2∴3.5n的2次方乘以m+2nm-2分之7mn的2次方+3mn=3.5×2²×(

再答：不客气啦

x^2-3x-1=0x^2-3x=12x^3-3x^2-11x+8=2x^3-6x^2+3x^2-11x+8=2x(x^2-3x)+3x^2-11x+8=2x+3x^2-11x+8=3x^2-9x+8

3^(x+1)*(2^x-3^x)*2^(x+1)=2^2+3^26^(x+1)*(2^x-3^x)=2^2+3^2x>0时,2^x-3^x

3x^2-x=19x^4+12x^3-2x^2-7x+2005=3x^2×(3x^2-x)+5x×(3x^2-x)+(3x^2-x)-6x+2005=3x^2+5x+1-6x+2005=3x^2-x+

4^x-3*2^(x+1)+8

楼主,题貌似错了吧,无法分解.我做过类似这道题可以分解的：x^4-2x3-27x2-44x+7用待定系数法就是要设原式=(x2+ax+1)(x2+bx+7)或=(x2+ax-1)(x2+bx-7)[因

求图片再问：再答：再答：谢了

(x^2-3x)(x+2)

(2x^4-4x^3+ax^2+7x+b)/(x^2+x-2)=2x^2-6x+a+2余(17-a)x+b-2a-4按已知17-a=0b-2a-4=0则a=17b=38故应该乘以2x^2-6x+19

（x-2)²-4(x+1)²=0平方差公式：[(x-2)+2(x+1)][(x-2)-2(x+1)]=03x(-x-4)=0x(x+4)=0x=0或x=-4

x^(6n)+x^(4n)*x^(5n)=x^(6n)+x^(9n)=[x^(3n)]^2+[x^(3n)]^3=2^2+2^3=4+8=12

(x-y)^2(x-y)^3*(y-x)^2*(y-x)^3=-(x-y)^2(x-y)^3*(x-y)^2*(x-y)^3=-(x-y)^10

x^2-2(k-3)x+k^2-4k-1=0两根x+y=2(k-3)xy=k^2-4k-1M=xy=(x+y+2)^2/4-5x+y+2=2倍根号下(xy+5)>=2+2倍根号下xyxy

设x^4-x^3+4x^2+3x+5=(x^2+ax+1)(x^2+bx+5)=x^4+(a+b)x^3+(ab+b)x^2+(5a+b)x+5根据对应项系数相等,得a+b=-1①ab+b=4②5a+

x^6-2x^4+6x^3+x^2-6x+9=(x^6-2x^4+x^2)+(6x^3-6x)+9=(x^3-x)^2+6(x^3-x)+9=(x^3-x+3)^2

y'=[(2x+1)²(3x+4)³]'=[(2x+1)²]'(3x+4)³+(2x+1)²[(3x+4)³]'=2(2x+1)(2x+1)