已知1x2分之1=1-2分之1
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/23 16:54:45
![已知1x2分之1=1-2分之1](/uploads/image/f/4204392-24-2.jpg?t=%E5%B7%B2%E7%9F%A51x2%E5%88%86%E4%B9%8B1%3D1-2%E5%88%86%E4%B9%8B1)
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
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
方程两边同时乘以x(x+1)(x-1)得:5(x-1)+3(x+1)=7x解得:x=2检验:x(x+1)(x-1)=6所以x=2是原分式方程的解
=2x/(x+2)(x-2)-(x+2)/(x+2)(x-2)=[2x-(x+2)]/(x+2)(x-2)=(2-x)/(x+2)(x-2)=-1/(x+2)
x1+x2=5;x1x2=1;(1)x1/x2+x2/x1=(x1²+x2²)/(x1x2)=((x1+x2)²-2x1x2)/(x1x2)=(25-2)/1=23;(2
已知一元二次方程x2-(根号3+1)x+根号3-1=0的两根为x1,x2则由韦达定理x1+x2=√3+1x1*x2=√3-1所以1/x1+1/x2=(x1+x2)/(x1*x2)=(√3+1)/(√3
1x2分之1--2x3分之1--3x4分之1.--2012x2013分之1=1-1/2-1/2+1/3-1/3+1/4-.-1/2012+1/2013=1/2013再问:注意1x2前面是负的再答:-1
两边乘以(x+3)(x-3)得12-2(x+3)=x-312-2x-6=x-3-3x=-9x=3检验:x=3是增根∴方程无解
兄弟,真的很简单,但是没有时间给你做你看下韦达定理,全部是基本应用,及两根之和、两根之积的关系,一下就出了.比如第一题,直接通分,第一空-5,第二个-3,第三个是(x1-x2)²=(x1+x
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)=
1/1*2+1/2*3+1/3*4+.+1/98*99+1/99*100=[1-1/2]+[1/2-1/3]+[1/3-1/4]+.+[1/98-1/99]+[1/99-1/100]=1-1/2+1/
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.(1/2006-1/2007)=1-1/2007=2006/2007
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)=
问题一规律:-n×n+1分之1=_n分之一+n+1分之1再问:求后面
x1+x2=-3/2x1x2=-21/x1+1/x2=(x1+x2)/x1x2=(-3/2)/(-2)=3/4x1²+x2²=(x1+x2)²-2x1x2=(-3/2)&
x1+x2=4x1x2=-1(x1+x2)^2/(1/x1+1/x2)=(x1+x2)^2*x1x2/(x1+x2)=x1x2*(x1+x2)=-4