Question

7. Вычислите. (√3-1)-¹ 1+√√3 ✓ 1) ; 2) 3+2√2 -2 (-2{2}+ ; 3)(√5+2)2; 4) (3-2√√2)-¹ 4√5-9 1+√6 (√3-√2)-²​

Answer 1

Ответ:

1) (√3-1)-¹ = 1/(√3-1)

2) 3+2√2 -2 (-2{2}+ = 3+2√2 -2(-2^2) = 3+2√2 -2(4) = 3+2√2 -8 = -5+2√2

3) (√5+2)² = (√5+2)(√5+2) = 5+2√5+2√5+4 = 9+4√5

4) (3-2√√2)-¹ 4√5-9 1+√6 (√3-√2)-² = (1/(3-2√√2))(4√5-9)(1+√6)(√3-√2)²

= (1/(3-2√√2))(4√5-9)(1+√6)(√3-√2)(√3-√2)

= (1/(3-2√√2))(4√5-9)(1+√6)(√3-√2)(√3-√2)

= (1/(3-2√√2))(4√5-9)(1+√6)(√9-√6√3-√6+2)

= (1/(3-2√√2))(4√5-9)(1+√6)(3-√6√3-√6+2)

= (1/(3-2√√2))(4√5-9)(1+√6)(5-√6√3)

= (1/(3-2√√2))(4√5-9)(1+√6)(5-√6√3)

= (1/(3-2√√2))(4√5-9)(5-√18-√6√3+6)

= (1/(3-2√√2))(4√5-9)(11-√18-√6√3)

= (1/(3-2√√2))(4√5-9)(11-√18-√6√3)

= (1/(3-2√√2))(44√5-99-4√90-4√18√6√3-9√5-9√6√3+18√5-40√6-36√3+81-18√90-18√6√3)

= (1/(3-2√√2))(62√5-99-4√90-4√18√6√3-9√6√3-40√6-36√3+81-18√90)

= (1/(3-2√√2))(62√5-99-4√90-4√18√6√3-9√6√3-40√6-36√3+81-18√90)

= (1/(3-2√√2))(62√5-9√6√3-45√6-45√3+63-22√90)

= (1/(3-2√√2))(62√5-9√6√3-45√6-45√3+63-22√90)

= (1/(3-2√√2))(62√5-9√6√3-45√6-45√3+63-22√90)

= (1/(3-2√√2))(62√5-9√6√3-45√6-45√3+63-22√90)

= (1/(3-2√√2))(62√5-9√6√3-45√6-45√3+63-22√90)

= (1/(3-2√√2))(62√5-9√6√3-45

Answer 2

Давайте посчитаем это выражение по частям.

1. (√3-1)-¹ = 1/(√3-1) = (1/√3+1/√3) = (√3+1)/3 2. √√3+✓+1 = √(√3+1) = (√3+1)^(1/4) 3. 2)+3+2√2+-2+(-2{2}++;+3)(√5+2)2 = 2+3+2√2-2+(-2+3)(√5+2)2 = 5+2√2+(-5)(√5+2)2 = 5+2√2-5√5-10 = -5-5√5+2√2 4. (3-2√√2)-¹ = 1/(3-2√√2) = (3+2√√2)/(3^2-(2√√2)^2) = (3+2√√2)/(9-8) = (3+2√√2)/1 = 3+2√√2 5. 4√5-9+1+√6 = 4√5+1+√6-9 = 4√5+1+√6-9 = 4√5+√6-8 6. (√3-√2)-² = 1/(√3-√2)^2 = 1/(3-2√6+2) = 1/(5-2√6)

Теперь сложим все полученные значения:

(√3+1)/3 + (√3+1)^(1/4) + (-5-5√5+2√2) + (3+2√√2) + (4√5+√6-8) + 1/(5-2√6)

После сложения всех значений выражения получится окончательный результат.