4AB & # 178; - 4A & # 178; B-B & # 179; process

4AB & # 178; - 4A & # 178; B-B & # 179; process

The original formula = - B (4a & # 178; - 4AB + B & # 178;)
=-b(2a-b)²
If the quadratic trinomial ka2 + 4A + 1 about a is a complete square, then the value of K may be?
Fast
If the quadratic trinomial ka2 + 4A + 1 about a is a complete square, then the value of K may be 4
4a^2+4a+1=(2a+1)^2
4a2 KAB + 9b2 is a complete square, then K=______ .
4a2 KAB + 9b2 = (2a) 2-kab + (3b) 2, ■ - KAB = ± 2 × 2A × 3b, the solution is k = ± 12
It is known that the square of 4A + KAB + 9b is a complete square, and the constant k = ()
Square of 4A + KAB + square of 9b
=（2a±3b)^2
K = 2 * 2 * 3 = 12 or - 2 * 2 * 3 = - 12
What is the value of K if 4A squared - K A + 1 is a complete square? A4 B-4 C2 D + 4
Let 4A square - K A + 1 be a complete square, then the value of K is D plus or minus 4
4A ^ 2-ka + 1 = (2a + 1) ^ 2 or (2a-1) ^ 2
So k = - 4 or K = 4
The square of 4A can be written as the square of 2A bracket, the constant is 1, so the more complete square can be written as the square of (2a ± 1), the decomposition can get k can be positive and negative 4, so choose D
It is known that the solution of the system of equations x + 2Y = 5mx − 2Y = 9m satisfies 3x + 2Y = 19, and the value of M is obtained
X + 2Y = 5m, 1 x − 2Y = 9m, 2, 1 + 2 get x = 7m, 1 - 2 get y = - m, 3 × 7m + 2 × (- M) = 19, 1
There is a series of monomials: - x, 2x ^, - 3x ^ 3, 4x ^ 4 Write the nth and the nth + 1 monomials
This question should be like this: - x, 2x ^ 2, - 3x ^ 3, 4x ^ 4
If so, the answers are attached
6X-16=4X 3X+1=15-x 12x-1=15x-7 73+2(X-4)=81 7(4-X)=9(X-4)
6x-16 = 4x out of 6x-4x = 16, x = 8.3x + 1 = 15-x, 3x + x = 15-1, x3.5.12x-1 = 15x-7, 12x-15x = - 7 + 1, x = 2.73 + 2 (x-4) = 81, 2x = 81-73-8, x8.7 (4-x) = 9 (x-4), x = 4
If the value of 2 / (2x * 2 + 3x + 7) is 1 / 4, find the value of 1 / (4x * 2 + 6x-1)
Because: the value of 2 / (2x * 2 + 3x + 7) is 1 / 4,
So 2x * 2 + 3x + 7 = 8, so 2x * 2 + 3x = 1, so 4x * 2 + 6x = 2, so 1 / (4x * 2 + 6x-1) = 1
Because 2 / (2x ^ 2 + 3x + 7) = 1 / 4
2x^2+3x+7=8
We get 2x ^ 2 + 3x = 1
1/(4x^2+6x-1)=1/[2(2x^2+3x)-1]=1/(2*1-1)=1
4X (m-2) - 3x (m-2), where x = 1.5, M = 6, decompose the factor first, and evaluate after calculation
4x（m-2）-3x（m-2）=（4x-3x）（m-2）=X（m-2）
Substituting x = 1.5, M = 6 into the above formula, the original formula = 1.5 × (6-2) = 1.5 × 4 = 6
First remove the brackets, the original formula = 4xm-8x-3xm + 6x
Merge congeners = xm-2x
Extract the common factor = x (m-2)
Band value: 1.5 * (6-2)
The final result: 6