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1/x+1+1/x+4=1/x+2+1/x+3怎么算?

解: 等式两边分别通分 (x+4+x+1)/[(x+1)(x+4)]=(x+3+x+2)/[(x+2)(x+3)] (2x+5)/(x²+5x+4)=(2x+5)/(x²+5x+6) 等式两边分子相同,而分母x²+5x+4恒≠x²+5x+6 因此只有2x+5=0 x=-2.5,代入分式方程检验,分母均不等于0,x=-2.5...

解:如图

1/(x+1)+1/(x+4)=1/(x+2)+1/(x+3) 1/(x+1) -1/(x+2) = 1/(x+3)- 1/(x+4) 1/[(x+1)(x+2)] = 1/[(x+3)(x+4)] (x+1)(x+2) =(x+3)(x+4) x^2+3x+2 = x^2+7x+12 4x=-10 x=-5/2

x+3/x+4-x+4/x+5=x+1/x+2-x+2/x+3 (x²+8x+15-x²-8x-16)/(x+4)(x+5)=(x²+4x+3-x²-4x-4)/(x+2)(x+3) -1/(x²+9x+20)=-1/(x²+5x+6) ∴x²+9x+20=x²+5x+6 9x-5x=6-20 4x=-16 x=-4 ∵x=-4原方程无意义, ∴原方程...

由于 (1-x)(1+x+x^2)=1-x^3 1+x+x^2+x^3+x^4=(1-x^3)/(1-x)+x^3*(1+x) =(1-x^3)/(1-x)+[x^3*(1+x)*(1-x)]/(1-x) =(1-x^3)/(1-x)+[x^3*(1-x^2)]/(1-x) =(1-x^3)/(1-x)+(x^3-x^5))/(1-x) =(1-x^3+x^3-x^5)/(1-x) =(1-x^5)/(1-x)

f(x)=(1/4)^x-(1/2)^x+1 =(1/2)^(2x)-(1/2)^x+1 =(1/2)^(2x)-(1/2)^x+1/4+3/4 =((1/2)^x -1/2)^2+3/4 当x=1时,f(x)最小值 3/4 当x=-3时 (1/2)^(-3)=2^3=8 当x=2时 (1/2)^2=1/4 所以 f(x)最大值:f(-3)=(8-1/2)^2+3/4=(15/2)^2+3/4=(225+3)/4=57...

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1/1*2+2/1*2*3+3/1*2*3*4+4/1*2*3*4*5+5/1*2*3*4*5*6+6/1*2*3*4*5*6*7 =1-1/1*2+1/1*2-1/1*2*3+1/1*2*3-1/1*2*3*4+1/1*2*3*4-1/1*2*3*4*5+1/1*2*3*4*5-1/1*2*3*4*5*6+61//1*2*3*4*5*6-1/1*2*3*4*5*6*7 =1-1/1*2*3*4*5*6*7 =1-1/5040 =5039/5040...

x=8,y=6

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