First, simplify and then evaluate the square of a + radical 1-2a + A, where a = 9 Thank you very much

First, simplify and then evaluate the square of a + radical 1-2a + A, where a = 9 Thank you very much

The original formula = absolute value of a + (1-A)
∵1-a＜0
The original formula = a + (A-1) = 2a-1
When a = 9
The original formula = 2a-1 = 2 × 9-1 = 17

First simplify and then evaluate: (1 - A + 1 / 1) - the square of a + 2A + 1 / 1, where a = root 3-1

The original formula = [1 - 1 / (a + 1)] - [A / (a A2 + 2A + 1)] = 1 - 1 / (a + 1) - A / (a + 1) mm2 = (a + 1) / (a + 1) 2 - (a + 1) / (a + 1) Ω = [(a + 1) 2 - (a + 1) - A] / (a + 1) mm2 = (a + 2A + 1-A) / (a + 1) mm2 = (a + 2A + 1-a-1-a) / (a + 1) mm2 = a

The square of A-1 fraction a minus the square of a minus the square of the root of a-2a + 1, where a = 2-radical 3, simplify and then evaluate

The square of A-1 minus the square of a minus the square of a-2a + 1
=(a-1)(a+1)/(a-1)-(a-1)²/[a(a-1)]
=a+1-(a-1)/a
=(a²+a-a+1)/a
=(a²+1)/a
=a+1/a
=2-√3+1/(2-√3)
=2-√3+2+√3
=4

First simplify, then evaluate: (a + b) (a-b) + 2A2, where a = 1, B = 2

The original formula = A2-B2 + 2A2 = 3a2-b2,
When a = 1, B = 2, the original formula = 3-4 = - 1

First simplify, then evaluate: X / x + 2 - (x ^ 2 + 2x + 1) / x + 2 △ x ^ 2-1 / X-1, where x = radical 3-2

: X / x + 2 - (x ^ 2 + 2x + 1) / x + 2 △ x ^ 2-1 / X-1, where x = radical 3-2 = x / (x + 2) - (x + 1) 2 / (x + 2) × (x-1) / (x + 1) (x-1) = x / (x + 2) - (x + 1) / (x + 2) = - 1 / (x + 2) = - 1 / (√ 3-2 + 2) = - √ 3 / 3

First simplify, then evaluate: (X-2) / (x ^ 2-1) / (2x + 2) / (x ^ 2 + 2x + 1) + 1 / (x-1), where x = radical 2 + 1

(x-2)/(x^2-1)÷(2x+2)/(x^2+2x+1)+1/(x-1)=(x-2)/(x-1)(x+1)÷2(x+1)/(x+1)²+1/(x-1)=(x-2)/(x-1)(x+1)*(x+1)/2+1/(x-1)=(x-2)/2(x-1)+1/(x-1)=[(x-2)+2]/2(x-1)=x/2(x-1)=(√2+1)/2(√2+1-1)=(2+√2)/4

It is known that cos (π / 6-A) = radical 3 / 3 Given cos (π / 6-A) = (radical 3) / 3, find cos [(5 π) / 6 + a] - Sin ^ 2 (a - π / 6) How to deal with the root 3 / 3 + sin ^ 2 (π / 6-A) and sin ^ 2 (π / 6-A) at present? Wait until 23:30

From the formula (Sina) ^ 2 + (COSA) ^ 2 = 1
Sin ^ 2 (π / 6-A) = 1-cos ^ 2 (π / 6-A) = 1 - (radical 3 / 3) ^ 2 = 1-1 / 3 = 2 / 3

Given cos (π / 6-A) = radical 3 / 3, find cos (5 π / 6 + a) - cos 2 (π / 3 + a) Mr. Zhao's solution

cos(5π/6+a)=cos[π-(π/6-a)]=-cos(π/6-a)=-√3/3cos²(π/3+a)=cos²[(π/2)-(π/6-a)]=sin²(π/6-a)=1-cos²(π/6-a)=1-1/3=2/3∴ cos(5π/6+a)-cos²(π/3+a)=-√3/3+2/3

Given cos (π - α) = 3 / 3, find cos (5 / 6 π + α) - Sin 2 (α - 6 π)

Set it as the best

If cos (π / 6-x) = radical 3 / 3, the square of COS (5 π / 6 + x) - sin (x - π / 6)

Cos (5 π / 6 + 6 + x) - Sin ^ 2 (x - π / 6 / 6) = cos [π (π / 6-x)] - [1-cos ^ 2 (x - π / 6)] = - cos (π / 6-x) - [1-cos ^ 2 (x - π / 6)] = - cos (π / 6-x) - [1-cos ^ 2 (π / 6-x)] because cos (π / 6-x) = √ 3 / 3, so the original formula = - √ 3 / 3 - [1 - (√3 / 3 / 3 / 1 - (√ 3 / 3 / 1 - [1 - (√ 3 / 3 / 3 / 1 - [1 - (√3 / 3 / 3 / 3 / 3 / 3 / 3 - [1 - [1 - [1 - [1 - 3) ^ 2) = - √ 3 / 3-1 + 1 / 3 =