\(\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^8\cdot5^7\left(2^3+5^3\right)}{2^5\cdot5^4\left(2^3+5^3\right)}\\ =\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\\ =2^3\cdot5^3\\ =8\cdot125\\ =1000\)

\(9,\left(\dfrac{1}{x^3+1}-\dfrac{1}{x+1}-\dfrac{1}{x^2-x+1}\right):\dfrac{1}{x^3+1}\left(x\ne1\right)\\ =\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}-\dfrac{1}{x^2-x+1}\right)\cdot\left(x^3+1\right)\\=\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\cdot\left(x^3+1\right)\\=\dfrac{1-x^2+x-1-x-1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\left(x^3+1\right)\\=\dfrac{\left(-x^2-1\right)\left(x^3+1\right)}{\left(x+1\right)\left(x^2-1+1\right)}\\ =-x^2-1\)

`0,75 + 0,4x=29/60`

`=> 3/4 +2/5x=29/60`

`=> 2/5 x= 29/60 - 3/4`

`=> 2/5 x=-4/15`

`=> x=-4/15 : 2/5`

`=>x=-2/3`

__

`23-0,6x=5/12`

`=> 23-3/5 x =5/12`

`=> 3/5x= 23-5/12`

`=> 3/5x=271/12`

`=> x= 271/12 : 3/5`

`=>x=1355/36`

\(1,\dfrac{x}{x+y}-\dfrac{y}{y-x}\left(xy\ne0\right)\\ =\dfrac{x}{x+y}+\dfrac{y}{x-y}\\ =\dfrac{x\left(x-y\right)}{x^2-y^2}+\dfrac{y\left(x+y\right)}{x^2-y^2}\\ =\dfrac{x^2-xy+xy+y^2}{x^2-y^2}\\ =\dfrac{x^2+y^2}{x^2-y^2}\\ 2,\dfrac{1}{x-1}-\dfrac{2}{x^2-1}\left(x\ge0;x\ne1\right)\\ =\dfrac{1}{x-1}-\dfrac{2}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x+1-2}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x+2}\)

\(3,\dfrac{x+2}{x^2+xy}-\dfrac{y-2}{xy+y^2}\left(x,y\ne0\right)\\ =\dfrac{x+2}{x\left(x+y\right)}-\dfrac{y-2}{y\left(x+y\right)}\\ =\dfrac{y\left(x+2\right)}{xy\left(x+y\right)}-\dfrac{x\left(y-2\right)}{xy\left(x+y\right)}\\ =\dfrac{xy+2y-xy+2x}{xy\left(x+y\right)}\\ =\dfrac{2y+2x}{xy\left(x+y\right)}\\=\dfrac{2\left(y+x\right)}{xy\left(x+y\right)}\\ =\dfrac{2}{xy}\)

\(4,\dfrac{1}{2x^2-3x}-\dfrac{1}{4x^2-9}\left(x\ne\pm\dfrac{3}{2}\right)\\ =\dfrac{1}{x\left(2x-3\right)}-\dfrac{1}{\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{2x+3}{x\left(2x-3\right)\left(2x+3\right)}-\dfrac{x}{x\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{2x+3-x}{x\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{x+3}{x\left(2x-3\right)\left(2x+3\right)}\)

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