√1-（8/17）²

√1-（8/17）²

Mathematical answer group for you to answer, I hope to help you
√[1-（8/17）²]
=√[（1 + 8/17）（1 - 8/17）]
=√[(25/17）（9/17）]
=√（15/17）²
=15/17
=√(1-64/289)
=√(225/289)
=15/17
√1-（8/17）²
=√(1-64/289）
=√(225/289）
=±15/17
(x-1) ² = 17 to find x
X = 1 + radical 17 or 1-radical 17
X & # 178; + 1 = 5. Find the size of X
Plus or minus 2
Given X & # 178; - 5x-2 = 0, find (X-2) &# 179; - (x-1) &# 178; + 1 =?
It is known that X & # 178; - 5x-2 = 0
x^2-5x=2
(X-2) & # 179; - (x-1) & # 178; + 1
=[(x-2)^3-(x-1)^2+1]/(x-2)
={(x-2)^3-[(x-1)^2-1]}/(x-2)
={(x-2)^3-[(x-1-1)(x-1+1)]}/(x-2)
={(x-2)^3-[(x-2)x]}/(x-2)
=(x-2)[(x-2)^2-x]/(x-2)
=(x-2)^2-x
=x^2-5x+4
=2+4
=6
(X-2) & # 179; - (x-1) & # 178; + 1 = (X-2) & # 178; - x = x & # 178; - 5x + 4
Because X & # 178; - 5x-2 = 0, X & # 178; - 5x + 4 = 6
Given X & # 178; - 5x-2007 = 0, find the value of (X-2) &# 179; - (x-1) &# 178; + 1 of (X-2)
(X-2) the (X-2) of the (X-2) [(X-2) [(x-1) [(x-1) [(x-1) [(x-1) [(x-1) [(x-1) [(x-1) & \\\\\\\\\\\35\\35\\\\\\\\\\\\\\\\\\\\\\\\\\\\x + 4 ∵ X & # 178; - 5x-2007 = 0 ∵ X & # 17
[(x-2)³-(x-1)²+1]/(x-2)
=[(x-2)³-(x-1+1)(x-1-1)]/(x-2)
=(x-2)²-x
=x²-5x+4
=x²-5x-2007+2011
=2011
[(x-2)³-(x-1)²+1]/(x-2)
=[(x-2)³-(x-1+1)(x-1-1)]/(x-2)
=[(x-2)³-x(x-2)]/(x-2)
=(x-2)²-x
=x²-5x+4
x²-5x-2007=0 x²-5x=2007
That is, X & # 178; - 5x + 4 = 2011
(x ^ 2-6) ^ 2-6 (x ^ 2-6) + 9 - > = [(X & # 178; - 6) - 3] &# 178?
I have written: take (X & # 178; - 6) as a whole, you can take it as T. the original formula = T & # 178; - 6T + 9 = (T-3) &# 178; that is = [(X & # 178; - 6) - 3] &# 178; & nbsp; ~ 523 will always answer for you, I wish you progress in your study ~ ~ ~ if you agree with my reply, Please click the [adopt as satisfactory answer] button in time ~ ~ the mobile phone questioner can click "satisfied" in the upper right corner of the client. ~ your adoption is my driving force ~ ~ ~ if there are any new questions, please ask me for help. The answer is not easy, please understand~~
(x^2-6)^2-6(x^2-6)+9
=(x²-6)²-2×(x²-6)×3+3²
=[(x²-6)-3]²
Complete square formula
(x²-6)²-2×(x²-6)×3+3²=[(x²-6)-3]²
a² -2 a b+b²=(a -b)²
It's based on the complete square formula.
(a-b)²=a²-2ab+b²
In this problem, a = (x ^ 2-6), B = 3
So, (x ^ 2-6) ^ 2-6 (x ^ 2-6) + 9 - > = [(X & # 178; - 6) - 3] &# 178;