A^2-14ab+49b^2=(a-7b)^2
49a^2+14ab+b^2=(7a-b)^2
4a^2-4ab+b^2=(2a-b)^2
36z^2+48zx+16^2=12z(3z-4x)+256
25a^2b^2+10ab+1=(5ab+1)^2
c^2-2ck^3+k^6=(c-k^3)^2
1)
![25^{x+1}-29*10^x+4^{x+1}=0](https://tex.z-dn.net/?f=25%5E%7Bx%2B1%7D-29%2A10%5Ex%2B4%5E%7Bx%2B1%7D%3D0)
![25*5^{2x}-29*5^x*2^x+4*2^{2x}=0](https://tex.z-dn.net/?f=25%2A5%5E%7B2x%7D-29%2A5%5Ex%2A2%5Ex%2B4%2A2%5E%7B2x%7D%3D0)
Делим все на
![2^{2x}](https://tex.z-dn.net/?f=2%5E%7B2x%7D)
![25* (\frac{5}{2} )^{2x}-29*( \frac{5}{2} )^x+4=0](https://tex.z-dn.net/?f=25%2A+%28%5Cfrac%7B5%7D%7B2%7D+%29%5E%7B2x%7D-29%2A%28+%5Cfrac%7B5%7D%7B2%7D+%29%5Ex%2B4%3D0)
Замена
![y=( \frac{5}{2} )^x\ \textgreater \ 0](https://tex.z-dn.net/?f=y%3D%28+%5Cfrac%7B5%7D%7B2%7D+%29%5Ex%5C+%5Ctextgreater+%5C+0)
при любом x.
25y^2 - 29y + 4 = 0
D = 29^2 - 4*25*4 = 841 - 400 = 441 = 21^2
y1 = (5/2)^x = (29 - 21)/50 = 8/50 = 4/25 = (5/2)^(-2); x1 = -2
y2 = (5/2)^x = (29 + 21)/50 = 1 = (5/2)^0; x2 = 0
2)
![\frac{7}{x+3}- \frac{12}{x-2} \leq 10](https://tex.z-dn.net/?f=+%5Cfrac%7B7%7D%7Bx%2B3%7D-++%5Cfrac%7B12%7D%7Bx-2%7D++%5Cleq+10)
![\frac{7(x-2) - 12(x+3) - 10(x-2)(x+3)}{(x-2)(x+3)} \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B7%28x-2%29+-+12%28x%2B3%29+-+10%28x-2%29%28x%2B3%29%7D%7B%28x-2%29%28x%2B3%29%7D++%5Cleq+0)
![\frac{7x-14-12x-36-10x^2-10x+60}{(x-2)(x+3)} \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B7x-14-12x-36-10x%5E2-10x%2B60%7D%7B%28x-2%29%28x%2B3%29%7D++%5Cleq+0)
Умножим обе части на -1, чтобы x^2 был с плюсом.
При этом поменяется знак неравенства. И приведем подобные.
![\frac{10x^2+15x-10}{(x-2)(x+3)} \geq 0](https://tex.z-dn.net/?f=%5Cfrac%7B10x%5E2%2B15x-10%7D%7B%28x-2%29%28x%2B3%29%7D++%5Cgeq+0)
Разделим все на 5
![\frac{2x^2+3x-2}{(x-2)(x+3)} \geq 0](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5E2%2B3x-2%7D%7B%28x-2%29%28x%2B3%29%7D+%5Cgeq+0)
Решим уравнение в числителе
D = 3^2 - 4*2(-2) = 9 + 16 = 25 = 5^2
x1 = (-3 - 5)/4 = -8/4 = -2
x2 = (-3 + 5)/4 = 2/4 = 1/2
![\frac{(2x-1)(x+2)}{(x+3)(x-2)} \geq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B%282x-1%29%28x%2B2%29%7D%7B%28x%2B3%29%28x-2%29%7D++%5Cgeq+0)
По методу интервалов
x ∈ (-oo; -3) U [-2; 1/2] U (2; +oo)
<span>{3 - (х - 2у) - 4у = 18,
{2х - 3у + 3 = 2(3х-у)
{3 - х + 2у - 4у = 18
{2х - 3у + 3 = 6х - 2у
{- х - 2у = 15
{- 4х - у = - 3
Второе умножаем на (-2)
{- х - 2у = 15
{8х + 2у = 6
Складываем
- х - 2у + 8х + 2у = 15 + 6
7х = 21
х = 21 : 7
х = 3
В первое уравнение </span>- х - 2у = 15 подставим х = 3.
- 3 - 2у= 15
2у = 15 + 3
2у = 18
у = 18 : 2
у = 9
Ответ: х = 3;
у = 9.
{х+5у=7|*(-3)
3х+2у=-5;
{-3х-15у=-21
3х+2у=-5;
{13у=-26;
у=-2
3х+2*(-2)=-5
3х-4=-5
3х=1
х=1/3.
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