x2 (y-3√x2)2=1怎么解
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An=(2n-1)x2^n=nx2^(n+1)-2^n,则Sn=[nx2^2x(2^n-1)/(2-1)]-[2x(2^n-1)/2-1]=(2^n-1)(4n-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
解题思路:由完全平方公式可求解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/rea
因为x2+y2=2y化为x2+(y-1)2=1,令x=sinθ,y=1+cosθ,所以2x+y=2sinθ+cosθ+1=5sin(θ+β)+1,其中tanβ=55.因为sin(θ+β)∈[-1,1]
原式=x+2y-x2+y=x2+3y,当x=4,y=19时,原式=42+3×19=1+1=2.
(X^2-y+1)(X^2+1)+X^2y+y-X^2=(X^2-y+1)(X^2+1)+(X^2+1)y-X^2=(X^2-y+y+1)(X^2+1)-X^2=(X^2+1)^2-x^2=(x^2+
y=x2+2x+1/(x2+2x+3)=(x+1)2/(x2+2x+3)当分母一定时,分子越小越好(x2+2x+3)=(x+1)2+2永远大于零当(x+1)2越小越好而X=-1时y=x2+2x+1/(
把x2-1x2+2x=y代入原方程,得:8y+3y=11.方程两边同乘以y得:3y2-11y+8=0.故选A.
y=2x/(3x²+1)=2/(3x+1/x)当x>0时3x+1/x>=2√3x*(1/x)=2√3此时0
x(x-1)-(x2-y)=-3x2-x-x2+y=-3-x+y=-3所以x-y=3所以原式=(x-y)2=9
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
解题思路:完全平方公式解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.
拥有界性法y大于等于-2小于1
y'=[(4x^3+2x)(x^2+2)/(x^4+x^2)-2xln(x^4+x^2)]/[x^2+2]^2=[(4x^3+2x)(x^2+2)-2x^3(x^2+1)ln(x^4+x^2)]/[(
就把(x2+1/x)=y代进去(2(x2+1)/x)+(6x/x2+1)=7化为2y+6/y=7解之得y=3/2或2(x2+1/x)=3/2时无解(x2+1/x)=2时x=1再问:用换元法把(x2+1
y=(x^2-2x-3)/2x^2+2x+1,x^2-2x-3=2yx^2+2yx+y(2y-1)x^2+(2y+2)x+y+3=0(2y+2)^2-4(2y-1)(y+3)>=0-4=
y=(x-1)(x-3)/(x-1)(2x+1)==>x不等于1y=(x-3)/(2x+1)==>y不等于-2/3y=1/2-7/(4x+2)==>y的值域(负无穷,-2/3)U(-2/3,1/2)U
再问:还是不清楚呀再问:y=(x-1)(x-3)/(x-1)(2x+1)==>x不等于1y=(x-3)/(2x+1)==>y不等于-2/3y=1/2-7/(4x+2)==>y的值域(负无穷,-2/3)
y={2(2x+1)^3-x6(2x+1)^2}/(2x+1)^6再问:过程呢????请写一下再答:这是一个公式,高等数学上册的,分母先平方,分子:分子求导乘以分母减去分子乘以分母求导
原式=2x2-2xy-6x2+9xy-2x2+2x2-xy+y2=-2x2+5xy+2y2,当x=-1,y=-2时,原式=-2+10+8=16.