Math Problem Statement

solve the promblems

Solve the given mathematical problems as shown in the image.

Solution

  1. Стойността на израза (2+3)24(31)(2 + \sqrt{3})^2 - 4(\sqrt{3} - 1): (2+3)2=4+43+3=7+43(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} 4(31)=4344(\sqrt{3} - 1) = 4\sqrt{3} - 4 7+43(434)=7+4=117 + 4\sqrt{3} - (4\sqrt{3} - 4) = 7 + 4 = 11 Отговор: А) 11.

  2. Допустими стойности на израза x+74x\frac{x+7}{\sqrt{4-x}}: 4x\sqrt{4-x} е дефинирана само когато 4x04-x \geq 0 или x4.x \leq 4. Отговор: Г) x<4x < 4.

  3. Корените на уравнението x2+9=13\sqrt{x^2 + 9} = \sqrt{13}: x2+9=13    x2=4    x=±2x^2 + 9 = 13 \implies x^2 = 4 \implies x = \pm 2 Отговор: А) 2-2 и 22.

  4. Изразът sin2(90α)+sin2α1\sin^2(90^\circ - \alpha) + \sin^2\alpha - 1: sin(90α)=cosα\sin(90^\circ - \alpha) = \cos\alpha cos2α+sin2α=1\cos^2\alpha + \sin^2\alpha = 1 11=01 - 1 = 0 Отговор: В) 0.

  5. Редицата {an}\{a_n\} е определена с равенствата a1=2a_1 = 2, an+1=2an1a_{n+1} = 2a_n - 1: a1=2,a2=221=3,a3=231=5a_1 = 2, \quad a_2 = 2 \cdot 2 - 1 = 3, \quad a_3 = 2 \cdot 3 - 1 = 5 2+3+5=102 + 3 + 5 = 10 Отговор: В) 10.

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Math Problem Analysis

Mathematical Concepts

Algebraic expressions
Inequalities
Square root equations
Trigonometric identities
Recursive sequences

Formulas

(a + b)^2 = a^2 + 2ab + b^2
Domain restrictions for square roots (x ≥ 0)
Pythagorean identity: sin^2(α) + cos^2(α) = 1

Theorems

Properties of quadratic equations
Trigonometric transformations
Recursive sequence rules

Suitable Grade Level

Grades 9-12