Question

va rog ajutor am nevoie de ele mâine ! ​

Answer 1

Explicație pas cu pas:

1) suma vârstelor:

M + I = 26

în urmă cu 4 ani:

M - 4 = 2(I - 4)

M = 2(I - 4) + 4

M = 2(I - 4 + 2)

M = 2(I - 2)

Maria nu poate să aibă în prezent 15 ani deoarece vârsta ei este un multiplu de 2.

I = 26 - M

M = 2(26 - M - 2)

M = 48 - 2M

3M = 48 => M = 16

Maria are 16 ani

I = 26 - 16 = 10

Irina are 10 ani

2)

a)

$E(x) = (1 + 3 \sqrt{2}x)(3 \sqrt{2}x - 1) - 3 {(2x - 1)}^{2} - (4x + 3)(x - 1) - {x}^{2} - 10x + 3$

$= 18 {x}^{2} - 1 - 3(4 {x}^{2} - 4x + 1) - (4 {x}^{2} - x - 3) - {x}^{2} - 10x + 3$

$= 18 {x}^{2} - 1 - 12 {x}^{2} + 12x - 3 - 4 {x}^{2} + x + 3 - {x}^{2} - 10x + 3$

$= {x}^{2} + 3x + 2 = (x + 1)(x + 2)$

b)

$\frac{1}{E(0)} + \frac{1}{E(1)} + \frac{1}{E(2)} + ... + \frac{1}{E(n)} = \frac{2022}{2023}$

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ... + \frac{1}{(n + 1)(n + 2)} = \frac{2022}{2023}$

$\left(\frac{1}{1} - \frac{1}{2} \right) + \left(\frac{1}{2} - \frac{1}{3} \right) + \left(\frac{1}{3} - \frac{1}{4} \right) + ... + \left(\frac{1}{n + 1} - \frac{1}{n + 2} \right) = \frac{2022}{2023}$

$\frac{1}{1} - \frac{1}{n + 2} = \frac{2022}{2023}$

$\frac{n + 2 - 1}{n + 2} = \frac{2022}{2023} < = > \frac{n + 1}{n + 2} = \frac{2022}{2023} \\ = > n = 2021$

3)

$f(x) = \frac{3}{2}x + 2 \\ g(x) = - \frac{x}{2} + 5$

a)

$f(4) + f(-4) = \frac{3}{2} \times 4 + 2 + \frac{3}{2} \times ( - 4) + 2 \\ = 6 + 2 - 6 + 2 = 4$

b) Gf ∩ Gg = {P}

$f(x) = g(x) => \frac{3x}{2} + 2 = - \frac{x}{2} + 5 \\ 3x + 4 = - x + 10 < = > 4x = 6 \\ = > x = \frac{3}{2} \\ f\left( \frac{3}{2} \right) = \frac{3}{2} \times \frac{3}{2} + 2 = \frac{9 + 8}{4} = \frac{17}{4} \\ => P\left( \frac{3}{2}; \frac{17}{4} \right)$

c) A(-2; -1), B(2; 5), M(a; b), M ∈ Oy a.î. MA + MB = minim

M ∈ Oy => a = 0

MA + MB = minim => M ∈ AB

Ecuația dreptei AB:

$A(x_{A}; y_{A}), B(x_{B}; y_{B})$

$\frac{y - y_{B}}{y_{A} - y_{B}} = \frac{x - x_{B}}{x_{A} - x_{B}}$

$\frac{y - 5}{-1 - 5} = \frac{x - 2}{-2 - 2} \\ \frac{y - 5}{-6} = \frac{x - 2}{-4}$

$2(y - 5) = 3(x - 2) \\ 2y - 10 = 3x - 6 \\ 2y = 3x + 4 \\ y = \frac{3}{2} x + 2$

$a = 0 => b = 2 => M(0; 2)$

4)

ΔABC, cu AB = 2 cm, AC = 3 cm, BC = 4 cm

M este mijlocul lui [AB] => AM ≡ MB

AB ÷ 2 = 2÷2 = 1 cm => AM = 1 cm

N ∈ [AC] a.î. ∢ABC = ∢ANM

=> ΔABC ~ ΔANM

$\frac{AB}{AN} = \frac{AC}{AM} = \frac{BC}{NM} = k$

$\frac{2}{AN} = \frac{3}{1} = \frac{4}{NM} = k \\ = > k = 3$

$AN = \frac{2}{3} \\ NM = \frac{4}{3}$

a)

$P_{(ANM)} = AN + NM + AM = \frac{2}{3} + \frac{4}{3} + 1 \\ = \frac{6}{3} + 1 = 2 + 1 = 3 \: cm$

b)

$\frac{A_{(ABC)}}{A_{(ANM)}} = {k}^{2} = {3}^{2} = 9$

$\frac{A_{(ABC)}}{A_{(ABC)} - A_{(ANM)}} = \frac{9}{9 - 1}$

$\frac{A_{(ABC)}}{A_{(MBCN)}} = \frac{9}{8} \\ = > A_{(MBCN)} = \frac{8}{9} \cdot A_{(ABC)}$