Question

ex 4 si 5
dau coroana.............​

Answer 1

Răspuns:

Explicație pas cu pas:

4.

Combinari de (a+b) luate cate a = (a + b)!/a!*(a + b - a)! = (a + b)!/a!b!

Combinari de (a+b) luate cate b = (a + b)!/b!*(a + b - b)! = (a + b)!/a!b!

(a + b)!/a!b! =  (a + b)!/a!b!

5.

Ecuatia dreptei y = ax + b

Calculam dreapta care trece prin A(3,3) si B(2,4)

3 = 3a + b

4 = 2a + b

b = 3 - 3a

4 = 2a + 3 - 3a

4 = 3 - a

a = 3 - 4

a = -1

b = 3 - 3*(-1) = 3 + 3 = 6

y = -x + 6

C(2m, 1-m) apartine dreptei

1- m = -2m + 6

2m - m = 6 - 1

m = 5

C(10, -4)

Answer 2

 

$\displaystyle\bf\\4)\\Folosim~formula:~~~~C_n^k=C_n^{n-k}\\\\Exemplu:~~~C_5^2=C_5^3=10\\\\Rezolvare:\\\\\boxed{\bf C_{a+b}^a=C_{a+b}^{a+b-a}=C_{a+b}^b}\\\\\\5)\\Folosim~formula:\\\\Punctele:~~A(x_{_A},~y_{_A}),~B(x_{_B},~y_{_B}),~C(x_{_C},~y_{_C})~sunt~coliniare~daca:\\\\\left|\begin{array}{ccc}\bf x_{_A}&\bf y_{_A}&\bf1\\\bf x_{_B}&\bf y_{_B}&\bf1\\\bf x_{_C}&\bf y_{_C}&\bf1\end{array}\right|=0$

.

$\displaystyle\bf\\Rezolvare:\\\\\left|\begin{array}{ccc}\bf3&\bf3&\bf1\\\bf2&\bf4&\bf1\\\bf2m&\bf 1-m&\bf1\end{array}\right|=0\\\\\\3\times4+2(1-m)+3\times2m-4\times2m-3\times2-3(1-m)=0\\\\12+2(1-m)+6m-8m-6-3(1-m)=0\\\\6m-8m+2(1-m)-3(1-m)+12-6=0\\\\-2m-(1-m)+6=0\\\\-2m+m-1+6=0\\\\-m+5=0~~\Big|\times(-1)\\\\m-5=0\\\\\boxed{\bf m=5}$