整体消参法,x=(1-t^2 1 t^2) (2t 1 t^2)
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x→∞lim(x^3-5x^2+1)^(1/3)-x=lim(x^3-5x^2+1)^(1/3)-(x^3)^(1/3)=lim[(x^3-5x^2+1)^(1/3)-(x^3)^(1/3)]*[(x
!已知x²-x-1=0x²-x=1;-x³+2x²+2013的值!=-x(x²-x)+x²+2013=x²-x+2013=1+20
是不是y=√(x-1)-x令t=√(x-1) 则x=t²+1,将t换成xy=√(x²)-x²+1∵x≥2∴y=-(x+1/2
选B,由x≥1时,f(x)=3^x-1确定x≥1时函数为增函数,又函数图象关于X=1对称,所以x
1/(x-1)-1/(x-2)=1/(x-3)-1/(x-4)1/(x-1)+1/(x-4)=1/(x-3)+1/(x-2)通分得(2x-5)/(x^2-5x+4)=(2x-5)/(x^2-5x+6)
4x+3y=11 2(2x+y)+y=112*5+y=11y=1将y=1带入2x+y=52x+1=52x=4x=2
(X-3)1/2-3Y=0(X-3)1/2=3Y2(X-3)-11=4*3Y-11=2YY=11/10X=18/5
定义域为Rf(-x)=ln(-x+√(x^2+1))-f(x)=-ln(x+√(x^2+1))=ln(1/x+√(x^2+1)),然后通分上下同乘x-√(x^2+1)得=ln(-x+√(x^2+1))
令y=x+1,则:f(y)=2(y-1)^2+5(y-1)+2=2y^2-4y+2+5y-5+2=2y^2+y-1所以:f(y)=2y^2+y-1即:f(x)=2x^2+x-1再问:这貌似不是整体带入
解题思路:把x+1/x=2代入到所求的代数式中即可得出结果解题过程:
或直接将1/√(1+x^2)dx凑成ln[x+√(1+x^2)]即可,满意请采纳,谢谢.
(2-x)/(x-3)+1/(3-x)=3(2-x)/(x-3)-1/(x-3)=3(2-x-1)/(x-3)=31-x=3x-94x=10x=2.5(5x-4)/(2x-4)=(2x+5)/(3x-
令t=√(x-2)(t≥0),那么x=t²+2于是y=t-1/2*(t²+2)=t-1/2*t²-1=-1/2*(t²-2t)-1=-1/2*(t-1)
x^2+x-1=0x^2=1-xx^3=x(1-x)=x-x^2=x-(1-x)=2x-1x^3-2x=-1x^3-2x+2011=2011-1=2010
再答:求采纳再问:思路会了就是最后有个解错了谢谢
(0.1x-0.02)/0.002-(0.1x+0.1)/0.05=3(50x-10)-(2x+2)=348x-12=348x=15x=15/48=0.3125
高三数学题函数f(x)=(1-t)ln(x-1)+x*x/2+(1-t)x+t*t/2+t,且t>11)求f(x)的单调区间解析:∵函数f(x)=(1-t)ln(x-1)+x*x/2+(1-t)x+t