1-x 2-x-8 3=1
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要使根号(x2+2x+4)-根号(x2-x+1)
设(x²-1)/(x²+2x)=t则8t+3/t=118t²-11t+3=0(8t-3)(t-1)=0解得t=3/8或t=11.t=3/8(x²-1)/(x
去分母得:x^2(y-1)+x(1-y)+y=0y=1时,上式无解y=1时,为二次式,须有delta>=0即(1-y)^2-4y(y-1)>=0(y-1)(3y+1)再问:x^2(y-1)+x(1-y
因为x^2=x+1,所以x^2-1=x两边平方有:(x^2-1)^2=x^4-2x^2+1=x^2则:x^4+1=3x^2所以x^2+1/x^2=(1+x^4)/x^2=3x^2/x^2=3
原式=(x²+3x+9)/(x-3)(x²+3x+9)-6x/x(x-3)(x+3)-(x-1)/2(x+3)=1/(x-3)-6/(x-3)(x+3)-(x-1)/2(x+3)=
你可以参见“韦达定理”方程两个根的积是1,说明他们互为倒数.x^2+1/x^2=(x+1/x)^2-2*x*1/x=(-5)²-2=23
X^2+2x-1=0x=[-2±√(2^2+4)]/2=-1±√2X^2+2x-1=(x+1-√2)(x+1+√2)
原式=(x+1)/(x-1)-x(x-2)/(x+1)(x-1)÷(x-2)(x+1)/(x+1)²=(x+1)/(x-1)-x/(x-1)=(x+1-x)/(x-1)=1/(x-1)请好评
令a=x2+x(a+1)(a+12)=42a2+13a+12=42a2+13a-30=0(a+15)(a-2)=0a=-15,a=2x2+x=-15x2+x+15=0无解x2+x=2x2+x-2=(x
x趋近无穷?如果是无穷,答案是1/2先有理化,然后再分子分母各除以x
令x²+x=t原方程变为t+1=6/tt²+t-6=0(t+3)(t-2)=0则t=2或-31)x²+x=2x²+x-2=0(x+2)(x-1)=0x=-2或x
x2+x+1=2/(x2+x)(X²+x)²+(x²+x)-2=0(x²+x+2)(x²+x-1)=0∴x²+x-1=0x=(-1±√5)/
(x²+x)(x²+x-2)=-1把(x²+x)看成整体(x²+x)[(x²+x)-2]=-1运用乘法分配率(x²+x)²-2(x
x²+x-1/(x²+x)=3/2两边同时乘以(x²+x)得:(x²+x)²-1=3(x²+x)/22(x²+x)²-3
5x²-3x-5=0△=3²-4×5×(-5)=109x=[﹣(﹣3)±√109]/5由原方程可得所求式子=(x+5)-1/(x+5)所求式子=(118±6√109)/25-25/
x³-x²-x-1=(x³-x)-(x+1)=x(x²-1)-(x+1)=x(x+1)(x-1)-(x+1)=(x²-x-1)(x+1)
令x=tant则dx=sec^2tdt于是∫dx/[x(x^2+1)]=∫sec^2t/[tantsec^2t]dt=∫dt/tant=∫(cost/sint)dt=∫(1/sint)dsint=ln
在(x2-1x)n展开式中通项为Tk+1=Ckn(x2)n−k(−1x)k=(-1)kCnkx2n-3k又x的一次项是第六项,即2n−3k=1k+1=6即n=8k=5故答案为8.
(x/x2-9)-(1/x2+6x+9)=x/(x+3)(x-3)-1/(x+3)²=[x(x+3)-(x-3)]/(x-3)(x+3)²=[x²+3x-x+3]/(x-
令a=x+1/xa²=x²+2+1/x²2(a²-2)-9a+14=0(2a-5)(a-2)=0x+1/x=5/22x²-5x+2=0(2x-1)(x