已知x的平方-3x-4=0,则代数式x的平方-x-4分之x
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3x^2+xy-2y^2=(3x-2y)(x+y)=0x=-2y/3,或x=-y但由于你要求的式子中分母里有X平方-Y平方x=-y舍去其他就代入吧
因为:x的平方-3x+1=0,所以:x的平方=3x-1.(将此式代入下式)因此,4x的平方-7x+4分之x的平方-2x+1=(3x-1-2x+1)/(12x-4-7x+4)=x/(5x)=1/5=0.
x^2+1/x^2=(x-1/x)^2+2=3^2+2=9+2=11
解;由3x²-2x-3=0得:3x²-2x=3所以:(-3x²+2x+5)/(6x²-4x-1)=[5-(3x²-2x)]/[2(3x²-2
解3x²-4x+6=03x²-4x=-6x²-3x+4+2x²-x+8——是不是少了个-x呢=3x²-4x+12=-6+12=6
5x²-5=3x所以5x²-2x-5=x5x²-2x=x+55x²-5=3xx²-1=3x/5所以原式=x+5-1/x=(x²+5x-1)/
2x^2-6x+3=02x-6+3/x=02x+3/x=6平方(2x+3/x)^2=364x^2+9/x^2+12=364x^2+9/x^2=24
x的平方-3x+1=0除以x得x-3+1/x=0x+1/x=3(x²+2x+1)/(4x²-7x+4)=(x+2+1/x)/(4x-7+4/x)=(3+2)/(4×3-7)=1
令X平方+4X=y,原式可变为y²+3y-18=0解得y=3或y=-6再解x²+4x=3或x²+4x=-6(无解)解得x=±根号7-2
x+1/x=3,x^2+1/x^2+2=9x^2+1/x^2=7(x-1/x)^2=X^2+1/x^2-2=7-2=5x-1/x=±√5x^2+y^2-4x-2y+5=0(x-2)^2+(y-1)^2
x^2-3x+1=0所以x-3+1/x=0x+1/x=3所以x^2+x^(-2)=7
x²+3x-1=0初一x得x-1/x=-3x²/(x^4-3x²+1)=1/[x²+1/x²-3]=1/[(x-1/x)²+2-3]=1/[
AnB=B,则B是A的子集B为空集,16-4a=4B不是空集,即a
4+3/x=0x=-3/4(2x-x^2)/(x^3+x^2)*(x^2+2x+1)/(x^2-4x+4)=[x(2-x)/x²(x+1)]*[(x+1)²/(x-2)²
x²+4x-2=0得x²+4x=2而3x²+12x+2004=3(x²+4x)+2004=3×2+2004=2010
x²-4=0x²=4原式=x(x²+2x+1)-x³-x²-x-7=x³+2x²+x-x³-x²-x-7=x&
4x前面是+还是-?再问:+再答:∵x²+4x-2=0∴x²+4x=2∴3x²+12x=6∴3x²+12x+2012=6+2012=2018
(x+1)/(x^2+4x+3)=0分母不为零∴x+1=0∴x=-1(x-3)/x^2+9/(3-x)=[(x-3)(3-x)+9x^2]/[x^2(3-x)]=[-9+6x-x^2+9x^2]/[x
(1-x)的平方+|y+2|=01-x=0y+2=0x=1y=-24xy-[2x的平方-3(x的平方-xy+x的平方)]=4xy-[2x^2-3x^2+3xy-3x^2]=4xy-[-4x^2+3xy