已知x-x分之1=4,求x2
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f(x-x分之一)=X2+x2分之一F(X-1/X)=X^2+1/X^2=(X-1/X)^2+2所以F(X)=X^2+2f(3)=9+2=11
你的题目应该是:2x/(x^2-1)=A/(x+1)+B/(x-1)对问题进行变形得:2x/[(x+1)(x-1)]=A/(x+1)+B/(x-1)对等式的右边进行通分整理得2x/[(x+1)(x-1
∵x2+4x+1=0,∴x+1x=-4,则原式=x2+1x2=(x+1x)2-2=16-2=14.
x+1/x=2(x+1/x)^2=4x^2+2+1/x^2=4x^2+1/x^2=2
x^2-3x+2=0(x-2)(x-1)=0x=2或x=1当x=2时x^2+1/x^2=2^2+1/2^2=4+1/4=17/4当x=1时x^2+1/x^2=1^2+1/1^2=1+1=2
已知:(x²+1)/x=x²/x+1/x=x+1/x所以:x²+1/x²=x²+1/x²+2-2=(x+1/x)²-2
x=7(x²-x+1)7x²-7x+7=x7x²+7=8x两边平方49x4+98x²+49=64x²两边减去49x²49x4+49x&sup
x2-3x+1=0,两边同时除以x得,x-3+1x=0,x+1x=3,两边平方得,x2+2+1x2=9,即x2+1x2=7,原式=1x2+3+1x2=17+3=110.
X2+Y2+8X+6Y+25=0x²+8x+16+y²+6y+9=0(x+4)²+(y+3)²=0∴x+4=0y+3=0x=-4y=-3X2+4XY+4Y2分之
x2+4x+1=0,可得x1=√3-2x2=-√3-2则x2+x-1为﹙3-5√3﹚或﹙5+5√3﹚
(x+1)(x-2)(x+3)(x-4)+16=0,(x²-x-2)(x²-x-12)+16=0(x²-x)²-14(x²-x)+24+16=0(x&
x+1/x=10x^2/(x^4+x^2+1)=1/(x^2+1+1/x^2)=1/[(x+1/x)^2-1]=1/(10^2-1)=1/99
x^(1/2)+x^(-1/2)=3求x^2+x^(-2)-(x^(3/2)+x^(-3/2))/2-3解:x^(1/2)+x^(-1/2)=3两边平方,得x+x^(-1)+2=9即x+x^(-1)=
解题思路:关键是理解所求的几何意义即可的了,,,,。解题过程:
14x^2-4x+1=0=>x^2+1=4x=>(x^2+1)^2=16x^2=>x^4+1=14x^2---------①x^2+(1/x^2)=(1/x^2)(x^4+1)----------②把
根号x-6分之a-x=根号x-6分之根号a-x则(a-x)^2=a-x解得x=a或x=a-1(1+x)倍根号x2-5x+4分之x2-1=(x-1)(x+1)/(x+1)√(x-4)(x-1)=√[(x
因为x又x分之1=3,即x+1/x=3所以x+1/x=7则x^4+1/x^4=47所以:(x^10+x^8+x^2+1)/(x^10+x^6+x^4+1)=
这是七年级下册的分式方程.1.去分母:两边同时乘X*(X-2)得X²+4-X²=a*(X-2)2.去括号,合并同类项得aX=2a+43.系数化为一得X=a分之2a+4因为方程无解,
x1+x2=4x1x2=-1(x1+x2)^2/(1/x1+1/x2)=(x1+x2)^2*x1x2/(x1+x2)=x1x2*(x1+x2)=-4