<span>25x^2-16=0</span>
<span>25x^2=16</span>
x^2=0,64
x=<u>+</u><span>0,8
<em>Ответ:</em><u>+</u> 0,8</span>
Уравнение прямой y = kx + b
Если прямая проходит через точки (2 ; 1) и (1 ; 0) , то подставим координаты этих точек в уравнение прямой
![\left \{ {{1=2k+b} \atop {0=k+b}} \right.\\\\ \left \{ {{b=-k} \atop {2k-k=1}} \right.\\\\ \left \{ {{k=1} \atop {b=-1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B1%3D2k%2Bb%7D+%5Catop+%7B0%3Dk%2Bb%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D-k%7D+%5Catop+%7B2k-k%3D1%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D1%7D+%5Catop+%7Bb%3D-1%7D%7D+%5Cright.+++)
Уравнение прямой : y =x - 1
Дальше объяснения аналогичные
2) (1 ; 2) (3 ; 4)
y = kx + b
2 = k + b 4 = 3k + b
![\left \{ {{2=k+b} \atop {4=3k+b}} \right.\\\\ \left \{ {{b=2-k} \atop {3k+2-k=4}} \right. \\\\ \left \{ {{b=2-k} \atop {2k=2}} \right. \\\\ \left \{ {{k=1} \atop {b=2-1=1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B2%3Dk%2Bb%7D+%5Catop+%7B4%3D3k%2Bb%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D2-k%7D+%5Catop+%7B3k%2B2-k%3D4%7D%7D+%5Cright.+%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D2-k%7D+%5Catop+%7B2k%3D2%7D%7D+%5Cright.+%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D1%7D+%5Catop+%7Bb%3D2-1%3D1%7D%7D+%5Cright.++)
y = k + b
3) (0 ; 2) (1 ; 0)
![\left \{ {{2=0*k+b} \atop {0=k+b}} \right.\\\\ \left \{ {{b=2} \atop {k=-2}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B2%3D0%2Ak%2Bb%7D+%5Catop+%7B0%3Dk%2Bb%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D2%7D+%5Catop+%7Bk%3D-2%7D%7D+%5Cright.++)
y = - 2x + 2
4) (- 1 ; 2) (2 ; - 1)
![\left \{ {{2=-k+b} \atop {-1=2k+b}} \right.\\\\ \left \{ {{b=k+2} \atop {-1=2k+k+2}} \right. \\\\ \left \{ {{b=k+2} \atop {3k=-3}} \right.\\\\ \left \{ {{k=-1} \atop {b=1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B2%3D-k%2Bb%7D+%5Catop+%7B-1%3D2k%2Bb%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3Dk%2B2%7D+%5Catop+%7B-1%3D2k%2Bk%2B2%7D%7D+%5Cright.+%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3Dk%2B2%7D+%5Catop+%7B3k%3D-3%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D-1%7D+%5Catop+%7Bb%3D1%7D%7D+%5Cright.+++)
y = - x + 1
5) (0 ; 0) ( - 3 ; - 3)
![\left \{ {{0=0*k + b} \atop {-3=-3k+b}} \right. \\\\ \left \{ {{b=0} \atop {-3k=-3}} \right.\\\\ \left \{ {{b=0} \atop {k=1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B0%3D0%2Ak+%2B+b%7D+%5Catop+%7B-3%3D-3k%2Bb%7D%7D+%5Cright.+%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D0%7D+%5Catop+%7B-3k%3D-3%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D0%7D+%5Catop+%7Bk%3D1%7D%7D+%5Cright.+++)
y = x
6) (1 ; - 2) (- 3 ; - 5)
![\left \{ {{- 2=k+b} \atop {-5=-3k+b}} \right.\\\\ \left \{ {{b=-k-2} \atop {-5=-3k-k-2}} \right.\\\\ \left \{ {{b=-k-2} \atop {-4k=-3}} \right.\\\\ \left \{ {{k=0,75} \atop {b=-2,75}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B-+2%3Dk%2Bb%7D+%5Catop+%7B-5%3D-3k%2Bb%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D-k-2%7D+%5Catop+%7B-5%3D-3k-k-2%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D-k-2%7D+%5Catop+%7B-4k%3D-3%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D0%2C75%7D+%5Catop+%7Bb%3D-2%2C75%7D%7D+%5Cright.++++)
y = 0,75x - 2,75
Log3(7 - 4*x) <= 3
ОДЗ:
по определению логарифма его аргумент должен быть строго больше нуля.
7 - 4*x > 0
4*x < 7
x < 1.75
Переходим к самому решению. 3 можно представить как log3(9), тогда
log3(7 - 4*x) <= log3(9)
Т.к. основание логарифмов больше 1, то соотношение между логарифмами сохраняется и для их аргументов, т.е.
7 - 4*x <= 27
4x >= -20
x >= -5
Подключаем ОДЗ и получаем ответ:
- 5 <= x < 1.75
Под б видимо пропустили х
196r² - p² = (14r)² - p² = (14r - p)(14r + p)
25x² - 289y² = (5x)² - (17y)² = (5x - 17y)(5x+17y)
a²b⁴ - 9c² = (ab²)² - (3c)² = (ab² - 3c)(ab² + 3c)
(m - 1)² - 121 = (m - 1)² - 11² = (m-1 - 11)(m-1 +11) = (m - 12)(m + 10)
c² - 2cm + m² = (c - m)² = (c-m)(c-m)
9 + 6c + c² = c² + 2*3*c + 3² = (c + 3)² = (c+3)(c+3)
81c² -36cm +4m² = (9c)² - 2*9c*2m + (2m)² = (9c - 2m)² = (9c-2m)(9c-2m)
125 + n³ = n³ + (5)³ = (n+5)(n² + 5n + 5²) = (n + 5)(n² + 5n + 25)
25c² + 10cm² + m = (5c)² + 2*5c*m² + m = 5c(5c + 2m²) + m = ???
если только не пропущена степень m:
25с² + 10сm² + m⁴ = (5c)² + 2*5c*m² + (m²)² =(5c+m²)² = (5c+m²)(5c+m²)