6-нұсқа Матем

$$\frac{\sqrt[]{6}}{3}$$

$$\frac{\sqrt[]{7}}{4}$$

$$\frac{\sqrt[]{6}}{2}$$

$$\frac{\sqrt[]{2}}{3}$$

$$\left[\frac{2}{3};1\right]$$

[-6;-3) υ (0;1]

(0;1]

(0; $$\left[\frac{2}{3}\right]$$

$$\frac{1}{\sqrt[]{\left(4x^2\ -1\right)\arcsin\left(-2x\right)}}$$

$$\frac{1}{\sqrt[]{\left(4x^2\ -1\right)\arcsin\left(2x\right)}}$$

$$\frac{1}{\sqrt[]{\left(1+4x^2\ \right)\arcsin\left(-2x\right)}}$$

$$\frac{1}{\sqrt[]{\left(1-4x^2\ \right)\arcsin\left(-2x\right)}}$$

(0;2) υ (4.2;5.25)

(1; 5.25)

(1; 4.2)

(0;1) υ (2; 4.2)

13

9

10

12

[0;+∞)

[1991;+∞)

[0;32]

[1990;1991)

$$\frac{x-3}{-2}=\frac{y+4}{-9}=\frac{z-5}{-8}$$

$$\frac{x-3}{2}=\frac{y+4}{9}=\frac{5-z}{8}$$

$$\frac{x-3}{2}=\frac{y+4}{4}=\frac{z-5}{3}$$

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

(-∞;0]

(-∞;1]

[0;1]

[1;+∞)

$$\sqrt[8]{2023}$$

$$\sqrt[5]{2023}$$

2023

$$\sqrt[4]{2023}$$

$$3x+4y\ge13\ \ \ \ \ \$$

$$2x-3y\ge-14\ \ \ \ \ \$$

$$5x+y<16\ \ \ \ \ \$$

$$3x+4y\ge13\ \ \ \ \ \$$

$$2x-3y\le14\ \ \ \ \ \$$

$$5x+y\ge16\ \ \ \ \ \$$

$$3x+4y\le13\ \ \ \ \ \$$ $$2x-3y\ge14\ \ \ \ \ \$$ $$5x+y\le16\ \ \ \ \ \$$

$$y\ge\frac{13}{4\ }-\frac{3}{4}x$$

$$y\le\frac{2}{3}x+\frac{14}{3}$$

$$y\le16\ -5x\ \ \ \ \$$