Given that Tana = 1 / 7, tanb = 1 / 3, and a, B ∈ (0, Pai / 4), then a + 2B =?

Given that Tana = 1 / 7, tanb = 1 / 3, and a, B ∈ (0, Pai / 4), then a + 2B =?

Given that Tana = 3, tanb = 2, a, B ∈ [π, 3 π / 2], then a + B is

tan(A+B)=(tanA+tanB)/(1-tanAtanB)=(3+2)/(1-3X2)=-1
And because a and B belong to (Wu, 3 Wu / 2)
So a + B belongs to (2, 3)
In the first and second quadrants. Tan (11 / 4) = - 1, so a + B = (11 / 4)

Calculation: 2cos 30 ° - (22 / 7-3.14) 0 power + root 12 + (1 / 3) - 1 power

2cos 30 ° - (22 / 7-3.14) to the power of 0 + √ 12 + (1 / 3) to the power of - 1
=The power of 2 * √ 3 / 2-1 + 2 √ 3 + 3
=√3-1+2√3+3
=3√3+2

√ 12 - (- & #189;) - 1 power Tan 60 ° + & #179; √ - 8 ＋| 3-2|
The absolute value of the difference between the root 12 minus - 1 / 2 minus 60 ° plus - 8 plus root 3 minus 2

√ 12 - (- & #189;) - 1 power Tan 60 ° + & #179; √ - 8 ＋| 3-2|
=2√3+2-√3/3-2+2-√3
=2√3/3+2

Why is the number of a & sup3; - 3A & sup2; B + 3AB & sup2; - 3B & sup3; 3?
Urgent! Please help

Because the degree of a polynomial is the degree of the highest degree term
The degree of a & sup 3; is 3
-The degree of 3A & sup2; B is 2 + 1 = 3
+The degree of 3AB & sup2; is 2 + 1 = 3
-The order of 3B & sup3; is 3
The highest item number is 3, so the number of a & sup3; - 3A & sup2; B + 3AB & sup2; - 3B & sup3; is 3

Given that a + B = 1, find the value of a & sup3; + B & sup3; + 3AB & sup2; + 3A & sup2; B

The original formula = (a + b) (A & sup2; - AB + B & sup2;) + 3AB (a + b)
=a²-ab+b²+3ab
=a²+2ab+b²
=(a+b)²
=1

Calculation: (2 / 3AB & sup2; - 2Ab) × 1 / 2Ab

：（2/3ab²-2ab）×1/2ab
=1/3a^2b^3-a^2b^2

It is known that the square of | A-2 | and (B + 1) are opposite to each other, and the value of 5A & sup2; b-2ab & sup2; + 3AB is obtained

Because they are opposite numbers, so | a - 2 | + (B + 1) &# 178; = 0, because the square of a number is greater than or equal to 0, and the absolute value of a number is greater than or equal to 0, the equation holds only when a - 2 = 0 and B + 1 = 0, so a = 2, B = - 1, so 5A & # 178; B - 2Ab & # 178; + 3AB = AB (5a - 2b

If B / a = 3 / 1, then the value of the fraction A & sup2; - 2ab-b & sup2; / A & sup2; - 3AB + B & sup2?

=a/b-2-b²/a²-3+b/a
=1/3-2-9-3+3
=1 / 3-11 this is (x), the correct is = A / b-2-b & sup2 / A & sup2; - 3 + B / A
=1/3-2-9-3+3
=1 / 3 + 11 that's right

(- 4) & sup2; + (three fourths) & ordm; = how much?

(- 4) &# 178; + (three fourths) &# 186; = 16 + 0 = 16