Решение<span>
y = -1/3x + 2
- x = 3y - 6
x = - 3y + 6
Если такое условие:
</span><span>y = -1/(3x) + 2
то решение будет таким:
</span>- 1 / (3x) = y - 2
-3x*(y - 2) = 1
x = - 1 / (3y - 6)
Из условия
![b_1+b_2+b_3=168](https://tex.z-dn.net/?f=b_1%2Bb_2%2Bb_3%3D168)
и
![b_4+b_5+b_6=21](https://tex.z-dn.net/?f=b_4%2Bb_5%2Bb_6%3D21)
. Найдем знаменатель, используя формулу n-го члена геометрической прогрессии.
![b_1+b_1q+b_1q^2=168~~\Rightarrow~~~ b_1(1+q+q^2)=168](https://tex.z-dn.net/?f=b_1%2Bb_1q%2Bb_1q%5E2%3D168~~%5CRightarrow~~~+b_1%281%2Bq%2Bq%5E2%29%3D168)
(*)
![b_1q^3+b_1q^4+b_1q^5=21~~~\Rightarrow~~~ b_1q^3(1+q+q^2)=21](https://tex.z-dn.net/?f=b_1q%5E3%2Bb_1q%5E4%2Bb_1q%5E5%3D21~~~%5CRightarrow~~~+b_1q%5E3%281%2Bq%2Bq%5E2%29%3D21)
(**)
Подставим теперь равенство (*) во второе равенство (**)
![168q^3=21\\\\ q= \sqrt[3]{ \dfrac{21}{168} } = \dfrac{1}{2}](https://tex.z-dn.net/?f=168q%5E3%3D21%5C%5C%5C%5C+q%3D+%5Csqrt%5B3%5D%7B+%5Cdfrac%7B21%7D%7B168%7D+%7D+%3D+%5Cdfrac%7B1%7D%7B2%7D+)
1. ( x - 2 ) ^ 2(x-3)+(x-2)(x-3)^2
=<span>(x-2)(x-3)(x-2+x-3)=</span><span>(x-2)(x-3)(2x-5)
2. 3m (m+2n)-2n(m+2n)^2=</span><span>(m+2n)(3m-2n(m+2n))=(m+2n)(3m-2mn-4n^2)
3. (p+3q)^2(p-q)-(p+3q)(p-q)^2</span>=<span>(p+3q)(p-q)(p+3q-(p-q))=</span><span>4q*(p+3q)(p-q)</span>
5(у-2x)=1/2z
5y-10x=0.5z
5y=0.5z+10x /5
y=0.1z=2x
------------------------------
5(у-2x)=1/2z
5y-10x=0.5z
-10x=0.5z-5y
10x=5y-0.5z /10
x=0.5y-0.05z
-----------------------------------
5(у-2x)=1/2z
5y-10x=0.5z
-0.5z=10x-5y
0.5z=5y-10x /0.5
z=10y-20x