Let's know the square of a + B - 4A + 6A + 13 = A. find the value of a + B

Let's know the square of a + B - 4A + 6A + 13 = A. find the value of a + B

A+B=9
Given that x2 + 2x + 5 is a factor of X4 + AX2 + B, find the value of a + B
Let X4 + AX2 + B = (x2 + 2x + 5) (x2 + MX + n) = X4 + (2 + m) X3 + (2m + N + 5) x2 + (5m + 2n) x + 5N. By comparing the coefficients of corresponding terms, we get 2 + M = 02m + N + 5 = a5m + 2n = 05n = B. the solutions are m = - 2, n = 5, a = 6, B = 25 ∥ a + B = 31
The roots of the equation 4x & # 178; - 3x = 0 are
4x²-3x=0
x(4x-3)=0
x1=0 x2=3/4
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Given the square of a + the square of B - 6a-8b + 25 = 0, find the value of (B / a) - (A / b)
a²+b²-6a-8b+25=0
a²-6a+9+b²-8b+16=0
(a-3)²+(b-4)²=0
a=3 b=4
a/b=3/4 b/a=4/3
b/a-a/b=4/3-3/4=(16-9)/12=7/12
a²-6a+9+b²-8b+16=0
(a-3)²+(b-4)²=0
So a = 3, B = 4
b/a-a/b=4/3-3/4=16/12-9/12=7/12
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The factorization of the second grade of junior high school has additional points
If x satisfies x ^ 5 + x ^ 4 + x = - 1, then x ^ 1998 + x ^ 1999 + +The value of x ^ 2004 is ()
A.2 B.1 C.3 D.5
2,(2003^2-4004*2003+2002*4008-2003*2004)/(2003^2-3005*2003-2003*2005+2005*3005)
3,[(7^4+64)(15^4+64)(23^4+64)(31^4+64)(39^4+64)]/[(3^4+64)(11^4+64)(19^4+64)(27^4+64)(35^4+64)]
4. Given that n is a positive integer and n ^ 4-16n ^ 2 + 100 is a prime number, find the value of n
For any integer (x, b) = x + 1, C + 1, x + 5, x + 1, C + 1
If it's all solved, I'll get extra points
If x satisfies x ^ 5 + x ^ 4 + x = - 1, then x ^ 1998 + x ^ 1999 + +The value of x ^ 2004 is (b) A.2 B.1 C.3 d.5 reason: from x ^ 5 + x ^ 4 + x = - 1, x ^ 5 + x ^ 4 + X + 1 = 0, x ^ 4 (x + 1) + (x + 1) = 0 (x ^ 4 + 1) (x + 1) = 0, only x + 1 = 0, x = - 1, x ^ 1998 + x ^ 1999 + +x^2004=1-1+1-1+1-1+1=12...
What is 3 (x ^ 3) ^ 2 * x ^ 3 - (3x ^ 3) ^ 3 + (5x) ^ 2 * x ^ 7?
What's the value of (- 2A) ^ 6 - (- 3A ^ 3) ^ 2 - [- (- 2A) ^ 2] ^ 3,
（1）
3(x^3)^2*x^3-(3x^3)^3+(5x)^2*x^7
=3x^9-27x^9+25x^9
=x^9
（2）
（-2a)^6-(-3a^3)^2-[-(-2a)^2]^3
=64a^6-9a^6+64a^6
=119a^6
Given the square of a + the square of b-6a-8b + 25 = 0, find the value of a + B
The square of a + the square of b-6a-8b + 25 = 0
The square of a + the square of b-6a-8b + 9 + 16 = 0
(a-3)²+(b-4)²=0
a=3,b=4
a+b=7
Square of a + square of B - 6a-8b + 25 = 0
The square of (A-3) + the square of (B-4) = 0
a=3 b=4
3a+4b===25
A = 3, B = 4, this is a complete square formula problem
Some of the problems are not clear. It's better to clarify the known conditions in the last figure
So, a = 3, B = 4, a + B = 7
Factorization,
1、 X-x;
2、 A & # 179; B & # 179; + A & # 178; B & # 178; - AB
3、 (a + b) & 178; - (a-b) & 178;
4、 ; - 178; ŧ
5、 12A & # 178; - 12ab + 3B & # 178;
6、 14xy-x & # 178; - 49y & # 178;
7、 (X-Y) & # - x + y
4x³-6x²=2x²(2x-3)a³b³+a²b²-ab=ab(a²b²+ab-1)（a+b)²-（a-b）²=(a+b+a-b)(a+b-a+b)=4ab-9a²+4=(2-3a)(2+3a)12a²-12ab+3b²=3(4a²-4ab+...
How to solve x ^ 2 + 5x + 7 = 3x + 11
x²+5x+7=3x+11
x²+5x-3x=11 -7
x²+2x=4
x²+2x+1=4+1
(x+1)²=5
∴x+1=±√5
x=-1±√5
x1=-1+√5,x2=-1-√5.
x^2+5x+7=3x+11
x^2+5x-3x+7-11=0
x^2+2x-4=0
x^2+2x+1-5=0
(x+1)^2-5=0
(x+1-√5)(x+1+√5)=0
X = - 1 + √ 5 or x = - 1 - √ 5
x^2+5x-3x+7-11=0
x^2+2x-4=0
(x-2)^2=0 x=2
Square of a + square of B + 4a-6a + 13 = 0, find A-B =?
4a-6a？ Ask: is the square of a + the square of B + 4a-6b + 13 = 0, find A-B =?