(5x+3)^2=(√13x)^2+[√13x+√(x^2+9)]^2,怎么解这个方程

xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 y=(3x^2-x√x+5√x-9)/√x 求导 已知1/√(x)+√(x)=√5,求√(x/x2+x+1)+√(x/x2-x+1)的值定义域 x>0,两边平方，x+2+1/x=5,x+1/x=3,x^2-3x+1=0,x^2+x+1-4x=0,x^2+x+1=4x,x^2-x+1-2x=0.x^2-x+1=2x,√[x/(x^2+x+1)]-√[x/(x^2-x+1)]=√[x/(4x)]-√[x/2x]=√（1/4）-√（1/2） 已知x*x-3x+1=0求√(x*x+1/x-2)=? 若x=√2,求代数式[(x^2-2x+4)/(√x^2-4x+4)]/[(x^3+8)/(x^2-4)]*[|6-x|/(x^2-5x-6)-[1/(x^2-x+1)]^-1的值 √(3)x+5=2x y=(3x^2-x√x+5√x-9)/√x 求导我算到4.5x^0.5-1+2x+4.5x^1.5 , 1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x+13x+14x+15x=550必须用解方程 已知x=13/(4+根号3),求(x^4-6x^3-2x^2+18x+23)/(x^2-8x+15) 的值 禁止复制粘贴：x=13/(4+√3)=4-√3√3=4-x3=x²-8x+16x^2-8x+13=0,所以x^2-8x+15=2；x^4-6x^3-2x^2+18x+23=x^2(x^2-8x+13)+2x^3-15x^2+18x+23=2x(x^2-8x+13)+x^2-8x+23=x^2-8x+2 *-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X || xy²/(x²y) - y×x²/(x²+x)=(x²-3x)/(x²-5x×2x) - 10/(x²-6x+9)=2a/(a²-4) - 1/(a-2)=2/(1-a ) - 1/(a-1)=x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 x+2x+3x+4x+5x+6x+7x+8x+9x=9x-8x-7x-6x-5x-4x-3x-2x-x.x等于多少? 帮个忙 求你们了 急用xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 填九宫格帮帮忙.x x 6 x x 7 x x 98 x x x 3 x 1 x x 9 x x 6 x 5 x 3 x x x 3 x x x x 1 8x x x 9 x 1 x x x2 1 x x x x 6 x x x 6 x 7 x 3 x x 1 x x 9 x 2 x x x 47 x x 8 x x 5 x x 求值域:f(x)=|x-1|+|x+3|+|x-2|f(x)=x+√(2x-1) 先化简在求值(x-1/3x-x+1/x)*x/x²-1,其中x=√2-3 (x-2)(x-3)+2(x+6)(x-5)-3(x²-7x+13),其中x=-1 (x-3/x-2)-(x-2/x-1)=(x-5/x4) -(x-4/x-3)