已知3x2 2y2=9x,求x2 y2的最大值-搜答案-智慧作业帮

1/10x4+3x2+1=x4-x3+(x3+3x2+x)-x+1=x4-x3+x(x2+3x+1)-x+1=x4-x3-x+1=x4-(x3+3x2+x)+3x2+1=x4-x(x2+3x+1)+3

∵x2+x-1=0，∴x2+x=1，2x3+3x2-x=2x（x2+x）+x2-x=2x+x2-x=x2+x=1．

因为x=√3+1所以x﹥0,x-1﹥0原式=√[x²/(1-2x+x²)]=√[x²/(x-1)²]=x/(x-1)=(√3+1)/(√3+1-1)=(√3+1

x^2+3x-2＝0,x(x+3)=2,∴(2x-4)╱(x^2+6x+9)÷（x-3）×(x^2-9)╱(2x-x^2)=2(x-2)/(x+3)^*1/(x-3)*(x+3)(x-3)/[x(2-

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

x+1/x=3(x+1/x)^2=3^2=9x^2+1/x^2+2=9x^2+1/x^2=7

f(x)=x²-x-5g(x)=1/3x³-5/2x²+4xg'(x)=x²-5x+4y=g'(x)/[f(x)+9]=(x²-5x+4)/(x

∵x2-9=0，∴x2=9，∴x2（x+1）-x（x2-1）-x-7=x3+x2-x3+x-x-7=x2-7，当x2=9时，原式=9-7=2．

由条件左右除以x,得x+1/x=3平方得：x2+1/x2+2=9所以x2+1/x2—2=5所求值为根号5

已知x2-3x+1=0，求x

x2-3x+1=0，两边同时除以x得，x-3+1x=0，x+1x=3，两边平方得，x2+2+1x2=9，即x2+1x2=7，原式=1x2+3+1x2=17+3=110．

x^2+3x+1=0方程两边同除以xx+3+1/x=0x+1/x=-3x^2+1/x^2=(x+1/x)^2-2=(-3)^2-2=9-2=7

x3-3x2-11x+8=x（x2-3x)-11x+8=x-11x+8=-10x+8

∵x²-5x+1=0∴x²=5x-1∴x³-4x²-4x-1=x.x²-4x²-4x-1=x(5x-1)-4(5x-1)-4x-1=5x&s

x2+4x+3=0,(x+3)(x+1)=0x=-3或者-1x=-3时2x3+9x2-2x-3的值=-2x27+81+6-3=30x=-1时2x3+9x2-2x-3的值=-2+9+2-3=6或者再问：

x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x

∵x^2+3x+5=7∴x^2+3x=23x^2+9x+2=3(x^2+3x)+2=3×2+2=8

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.

x2/x4+x2+1上下除以x^2=1/(x^2+1+1/x^2)=1/[(x+1/x)^2-1]=1/(9-1)=1/8

你好x²+3x-1=0两边同除以xx+3-1/x=0x-1/x=-3两边平方得x²+1/x²-2=9x²+1/x²=11x²+1/x

X2-3X-1=0则X-3-1/X=0则X-1/X=3则(X-1/X)²=3²=9则X²-2+1/X²=9则X²+1/X²=9+2=11