1/3√9x+2√x/4-x√1/x
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1/3√9x+2√x/4-x√1/x1/3√9x+2√x/4-x√1/x1/3√9x+2√x/4-x√1/x=1/3*3√x+2*1/2*√x-√(x²*1/x)=√x+√x-√x=√x
1/3√9x+2√x/4-x√1/x
1/3√9x+2√x/4-x√1/x
1/3√9x+2√x/4-x√1/x
=1/3*3√x+2*1/2*√x-√(x²*1/x)
=√x+√x-√x
=√x
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
化简:[√(x^2-6x+9)/x^2-x-12]*(x^3-16x)/(x^2-3x)-1/(x+3) {x>3}
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(4)2/3X√9x-(x²√1/x-6x√x/4)
2/3x√9x-(x^2√1/x-6x√4/x)
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求一个数独答案X X X 9 X X X 8 2 X 6 3 X X 1 4 X 99 X 8 X X X X X XX X X 6 7 X 3 X XX 4 6 X 5 X 2 9 XX X 7 X 2 3 X X XX X X X X X 7 X 17 X 4 3 X X 6 2 X6 3 X X X 7 X 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
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
lim((√1+x^2)+x)/((√x^3+x)-x)x→+无穷大
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已知x*x-3x+1=0求√(x*x+1/x-2)=?
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如下9*9宫图如何填x x x 1 x x x 2 9x x 5 x x x x x 4x 6 8 x x x x x xx x x 7 x x 5 x xx 2 x x 6 x x 8 xx x 3 x x 9 x x xx x x x 5 x 1 6 x 4 x x x 3 x x x x 7 x x x x 2 x x x
(1) 2/3√9x+6√x/4-2x√1/x
1/3√9x+2√x/4-x√1/x
2√(9x)/3+6√(x/4)-2x√(1/x)
化简:2√(9x)/3+6√(x/4)-2x√(1/x)