0,04х^2-у^2 При х=-8;у=0,6...

karinasamoylova1999 @karinasamoylova1999

August 2022 1 1 Report

1)упростить:5t^2/1-t^2:(1-1/1-t)
2)Выполнить действие:(4е/2-е-е):(е+2/е-2)
3)Представить в виде дроби:у+2/у^-2у+1*3у-3/у^2-4
4)Упростить и вычислить:0,2х-у/0,04х^2-у^2
При х=-8;у=0,6

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NNNLLL54

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$1)\; \; \frac{5t^2}{1-t^2}:(1-\frac{1}{1-t})=\frac{5t^2}{(1-t)(1+t)}:\frac{1-t-1}{1-t}=\frac{5t^2\, \cdot \, (1-t)}{(1-t)(1+t)\, \cdot \, (-t)}=-\frac{5t}{t+1}$

$2)\; \; (\frac{4e}{2-e}-e):(\frac{e+2}{e-2})=\frac{4e-2e+e^2}{2-e}\cdot \frac{e-2}{e+2}=\frac{e^2+2e}{-(e-2)}\cdot \frac{e-2}{e+2}=\frac{e(e+2)}{-(e+2)}=-e$

$3)\; \; \frac{y+2}{y^2-2y+1}\cdot \frac{3y-3}{y^2-4}=\frac{y+2}{(y-1)^2}\cdot \frac{3(y-1)}{(y-2)(y+2)}=\frac{3}{(y-1)(y-2)}$

$4)\; \; x=-8\; ,\; \; y=0,6\\\\\frac{0,2x-y}{0,04x^2-y^2}=\frac{0,2x-y}{(0,2x-y)(0,2x+y)}=\frac{1}{0,2x+y}=\frac{1}{-1,6+0,6}=\frac{1}{-1}=-1$

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