若y=lg[X^2 [k 3]x 4]

（x+2y）(x—y)=2xy整理得x²-xy-2y²=0因为xy均>0所以等号两边同时除以xy设x/y=t整理得t-1-2/t=0剩下的你自己接就行了~

把x=-1代入原方程得到-2-k3--1-3k2=1，去分母得：-4-2k+3+9k=6移项、合并同类项得：7k=7解得：k=1．故填：1．

∵lg（x-y）+lg（x+2y）=lg2+lgx+lgy，∴lg（x-y）（x+2y）=lg2xy．∴（x-y）（x+2y）=2xy，即（x-2y）（x+y）=0．再由x、y都是正数可得x+y≠0，

(x-y)(x+3y)=2^2*xyx^2+2xy-3y^2=4xyx^2-2xy-3y^2=0(x+y)(x-3y)=0x=-y,x=3y由定义域x>0,y>0x=-y不成立x=3yx/y=3

lg(x+2y)+lg(x-y)=lg2+lgx+lgylg(x+2y)(x-y)=lg2xy(x+2y)(x-y)=2xyx^2+xy-2y^2=2xyx^2-xy-2y^2=0(x-2y)(x+y

s=lgx^n+lg(x^(n-1)y)+lg(x^(n-2)y^2))+.+lgy^n={(lgx^n+lg(x^(n-1)y)+lg(x^(n-2)y^2))+.+lgy^n)+(lgx^n+lg

2lg(x-2y)=lgy+lgx即(x-2y)^2=xy(x-y)(x-4y)=0又x>2y>0故x=4y所以log2(x/y)=2

2lg(X-2Y)=lgX+lgYlg(X-2Y)^2=lgXY(X-2Y)^2=XYX^2+4Y^2-4XY-XY=0X^2-5XY+4Y^2=0(X/Y)^2-5(X/Y)+4=0(X/Y-4)(

有题可知：x+2y>0；x-4y>0；x>0；y>0可得：x/y>4原式化简为：lg(x+2y)(x-4y)=lg2xy所以：(x+2y)(x-4y)=2xy化简：x^2-4xy-8y^2=0同除y^

. -y)+lg(x+2y)=lg[(x-y)(x+2y)]lg^2+lgx+lgy=lg(2xy)所以：(x-y)(x+2y)=2xyx^2+2xy-xy-2y^2=2xyx^2-2y^2

(x-y)(x+2y)=2xyx^2+2xy-xy-2y^2=2xyx^2-2y^2=xyx/y-2y/x=1设a=x/y则有a^2-a-2=0(a-2)(a+1)=0a=2,-1.根据题意取正所以：

lg(x-y)+lg(x-2y)=lg2+lgx+lgylg(x-y)(x-2y)=lg2xyx^2-2xy-xy+2y^2=2xyx^2-5xy+2y^2=0(x/y)^2-5(x/y)+2=0x/

lg(x+y)+lg(2x+3y)=lg12+lgx+lgylg[(x+y)(2x+3y)]=lg(12xy)(x+y)(2x+3y)=12xy2x^2+5xy+3y^2=12xy=02x^2-7xy

∵h(x)＝2x−kx+k3∴h'（x）=2+kx2因为函数h（x）在（1，+∞）上是增函数，所以h'（x）=2+kx2≥0在（1，+∞）上恒成立即k≥-2x2在（1，+∞）上恒成立∴k≥-2故答案为

答：lg(x-3y)^2=lg4xy(x-3y)^2=4xyx^2-6xy+9y^2=4xyx^2-10xy+9y^2=0(x-y)(x-9y)=0x=yorx=9yy/x=1ory/x=1/9

2lg(x-3y)=lgx+lg(4y)lg(x-3y)²=lg(4xy)(x-3y)²=4xyx²-6xy+9y²-4xy=0x²-10xy+9y&

lg(x+y)+lg(2x+3y)-lg3=lg4+lgx+lgylg(x+y)+lg(2x+3y)=lg3+lg4+lgx+lgylg[(x+y)*(2x+3y)]=lg[3*4*xy)(x+y)(

lg√x=1/2lgxlg（y/10）∧2=2lg（y/10）=2(lgv-1)lg√x-lg（y/10）∧2=1/2a-2b+2

解方程得：x=3-4k，则3-4k＜0，解得：k＞34．

由lg(x-y)+lg(7x-8y)=2,得lg(x-y)(7x-8y)=2(x-y)(7x-8y)=100(1)由2^(-x+3y)=4,得-x+3y=2x=3y-2(2)把（2）代入（1）得2(y