How to fill in the brackets of X & # 178; + 1 / X & # 178; = (x + 1 / x) & # 178; - () = (x-1 / x) & # 178; + ()?

How to fill in the brackets of X & # 178; + 1 / X & # 178; = (x + 1 / x) & # 178; - () = (x-1 / x) & # 178; + ()?

From the complete square formula, we get: (a + b) &# 178; = A & # 178; + B & # 178; + 2Ab, we get: A & # 178; + B & # 178; = (a + b) &# 178; - 2Ab (a-b) &# 178; = a & # 178; + B & # 178; - 2Ab, we get: A & # 178; + B & # 178; = (a-b) &# # 178; + 2Ab, so: X & #

(1) Given x ^ 2-5x + 1 = 0, find the value of x ^ 4 = 1 / x ^ 4? (2) given x ^ 2 + 9y ^ 2-8x + 6y + 17 = 0, find the value of 1 / X-1 / y =?

"=" should be + right
x^4+1/x^4=(x+1/x)^2-2
Let x + 1 / x = t
We get x ^ 2-tx + 1 = 0
Compare x ^ 2-5x + 1 = 0
t=5
So x ^ 4 + 1 / x ^ 4 = (x + 1 / x) ^ 2-2 = 5 ^ 2-2 = 23
2、x^2+9y^2-8x+6y+17=(x^2-8x+16)+(9y^2+6y+1)
=(x-4)^2+(3y+1)^2=0
x=4,y=-1/3
So 1 / X-1 / y = 13 / 4

If the proposition ax ^ 2_ 2ax_ 3> If 0 is not true, then the value range of real number is

If the proposition ax ^ 2-2ax-3 > 0 is not true, then ax ^ 2-2ax-3 ≤ 0 is true, and its range is good. When a = 0 - 3 ≤ 0 2. A ＞ 0, the discriminant 4A & # 178; + 12a ≥ 0 can get the range of X; a ＜ 0 4A & # 178; + 12a ≥ 0, a ≤ - 3 can get the range of X; a ∈ (- 3

If proposition ax & # 178; - 2ax-3 does not hold true and is a true proposition, then the value range of real number a is

Hello
The proposition should be "ax ^ 2-2ax-3 > 0, for all x ∈ r does not hold" is true proposition. 1. When a = 0, the proposition is true proposition. 2. When a > 0, the proposition is not true proposition. 3