А)
6x+10y-6x-12y=-2y
б)
8a^3-8ba^2+48ab^2-5ab^2-30ab^2+15b^3+2ba^2-12ab^2
=8a^3+15b^3+ab^2-6ba^2
(3x² -7x+2)/(2-6x) =3(x-1/3)(x-2)/(-6(x-1/3)) = (2-x)/2 . * * * x≠1/3 * * *
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(x+40)/(x³-16x) :((x-4)/(3x² +11x-4) - 16/(16-x²)) =
(x+40)/x(x²-16) :( (x-4)/3(x+4)(x-1/3) +16/(x² -16))=
(x+40)/x(x²-16) :( (x-4)/(x+4)(3x-1) +16/(x² -16))=
(x+40)/x(x²-16) :( ((x-4)² +16(3x-1)) /(x²-16)(3x-1))=
(x+40)/x(x²-16) :( (x²-8x+16 +48x-16) / (x²-16)(3x-1)) =
x+40)/x(x²-16) :( (x² +40x) / (x²-16)(3x-1))
(x+40)/x(x²-16) *(x²-16)(3x-1)/(x²+40) = (3x-1)/x * * * x≠<span>±4 * * *</span>
Приведем все к одному виду:
8=√64
3√7=√(7*9)=√63
62<63<64<65
√62<√63<√64<√65
А значит, самое меньшее значение √62.
1. 6ab⁵ = -7
1) 3*6ab⁵ = 18ab⁵ = 3*(-7) = -21
2) (6ab⁵)² = 36a²b¹⁰ = (-7)² = 49; 36a²b¹⁰:6 = 6a²b¹⁰ = 49÷6 = ![\displaystyle \frac{48+1}6 =\bold{8\frac16 }](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B48%2B1%7D6%20%3D%5Cbold%7B8%5Cfrac16%20%7D%20)
2.
![\displaystyle 128x^2y^3\cdot (\frac{-1}4 xy^3)^3=128x^2y^3\cdot (\frac{-1}{4^3} x^3y^9)=\frac{-2\cdot 64}{64} x^{2+3} y^{3+9} =\bold{-2x^5y^{12}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20128x%5E2y%5E3%5Ccdot%20%28%5Cfrac%7B-1%7D4%20xy%5E3%29%5E3%3D128x%5E2y%5E3%5Ccdot%20%28%5Cfrac%7B-1%7D%7B4%5E3%7D%20x%5E3y%5E9%29%3D%5Cfrac%7B-2%5Ccdot%2064%7D%7B64%7D%20x%5E%7B2%2B3%7D%20y%5E%7B3%2B9%7D%20%3D%5Cbold%7B-2x%5E5y%5E%7B12%7D%7D%20)
-х+2у=4
7х-3у=5
х=2у-4
7(2у-4)-3у=5
14у-28-3у=5
11у=33
у=3
х=2
Ответ: (2;3)