Разберем каждую часть отдельно:
1)а⁵b³-4a³b=a³b(a²b²-4)=a³b((ab)²-4)=a³b(ab-2)(ab+2)
2)a²b-2a³=a²(b-2a)
Ceйчас разделим:
![\frac{a^3b(ab-2)(ab+2)}{a^2(b-2a)}=\frac{ab(ab-2)(ab+2)}{b-2a}= \frac{ab(a^2b^2-4)}{b-2a}=\frac{a^3b^3-4ab}{b-2a}=\frac{((ab)^3-4ab)}{b-2a}=](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%5E3b%28ab-2%29%28ab%2B2%29%7D%7Ba%5E2%28b-2a%29%7D%3D%5Cfrac%7Bab%28ab-2%29%28ab%2B2%29%7D%7Bb-2a%7D%3D+%5Cfrac%7Bab%28a%5E2b%5E2-4%29%7D%7Bb-2a%7D%3D%5Cfrac%7Ba%5E3b%5E3-4ab%7D%7Bb-2a%7D%3D%5Cfrac%7B%28%28ab%29%5E3-4ab%29%7D%7Bb-2a%7D%3D)
![\frac{( \sqrt[3]{64})^3-4* \sqrt[3]{64})}{ \sqrt[3]{2}-2 \sqrt[3]{32}}= \frac{64-16}{ \sqrt[3]{2}- \sqrt[3]{256}}= \frac{48}{ \sqrt[3]{2}(1- \sqrt[3]{128})}](https://tex.z-dn.net/?f=+%5Cfrac%7B%28+%5Csqrt%5B3%5D%7B64%7D%29%5E3-4%2A+%5Csqrt%5B3%5D%7B64%7D%29%7D%7B+%5Csqrt%5B3%5D%7B2%7D-2+%5Csqrt%5B3%5D%7B32%7D%7D%3D+%5Cfrac%7B64-16%7D%7B+%5Csqrt%5B3%5D%7B2%7D-++%5Csqrt%5B3%5D%7B256%7D%7D%3D+%5Cfrac%7B48%7D%7B+%5Csqrt%5B3%5D%7B2%7D%281-+%5Csqrt%5B3%5D%7B128%7D%29%7D)
-3x2 + 8x + 3 = 0
D = b2 - 4ac
D = 64 + 36 = 100 = 10^2
x1,2 = -b ± √D/2a
x1 = -8 + 10/-6= - 2/6 = - 1/3
x2 = -8 - 10/-6 = 18/6 = 3
Ответ: x1 = - 1/3 ; x2 = 3
-1_2/3(0.6x-6)-2/3(9x+1.5) = -5/3(3/5 x - 6) - 2/3(9x+1.5) = -x+10-6x-1 = -7x+9
при x=-2,1
-7 * (-2.1) + 9 = 14.7+9 = 23.7
0.4t(t3-q)+0.7q(t3-q)=(t3-q)(0.4t+0.7q)
777777777777777777777777777777777777777, см. прикрепленное изображение