If (3a + 2b) x2 + ax + B = 0 has a unique solution, then x is equal to () A. -32B. 32C. 23D. -23

If (3a + 2b) x2 + ax + B = 0 has a unique solution, then x is equal to () A. -32B. 32C. 23D. -23

According to the meaning of the question: 3A + 2B = 0, then Ba = - 32, the original equation is ax + B = 0, the solution is: x = - BA = 32
The same solution of an equation 1 / 3A + X / 3 = 0 and 4a-4y-4 = 0 about X
(1) Finding the value of a
（2）（x+y)+(x+y)²+(x+y)³+… +(x+y)2009
If it is 1 / (3a), then a has no solution. If it is a / 3, the solution of equation 1 is x = - A, and the solution of equation 2 is y = A-1. From the problem, we know that - a = A-1, a = 1 / 2 (2) x = - 1 / 2, y = - 1 / 2, x + y = - 1 (x + y) + (x + y) & # 178; + (x + y) & # 179; + +(x + y) 2009 is a summation problem of equal ratio sequence
(1) ∵ equation 1 / 3A + X / 3 = 0 and 4a-4y-4 = 0 have the same solution
∴X=Y
1/3a+x/3=0
The solution is a = - X
Substituting 4a-4y-4 = 0, - 4x-4y-4 = 0
After simplification, we can get: x + y + 1 = 0
x+y=-1
And ∵ x = y
∴x=y=-0.5
∴a=-x=0.5
(2) The original formula = - 1 + 1-1 + 1 - -1
=-1
One third times a + one third of x = 0 and 4a-4y-4 = 0 are the same solution of a linear equation of one variable with respect to X to find the value of A
A equals half
1 / 3 * a + X / 3 = 0 multiply by 3 A + x = 0 a = - x
About 4a-4y-4 = 0 divide by 4 a-y-1 = 0 a = y + 1
-X = y + 1, y + 1 + x = 0, is it true that x is written as y, because it is a linear equation with one variable
So x + 1 + x = 0, x = - 1 / 2
5y-4 ﹣ 2y-4 + 1 ﹣ 2 = 2Y + 5 ﹣ 3y-6
(5y-4)÷(2y-4)+1÷2=(2y+5)÷(3y-6)
(5y-4)÷2(y-2)+1÷2=(2y+5)÷3(y-2)
6(5y-4)+3(y-2)=2(2y+5)
30y-24+3y-6=4y+10
29y=40
y=40/29
If the circle x2 + Y2 + 2x-4y + 1 = 0 is symmetric with respect to the line 2ax-by + 2 = 0, (a, B belong to R), then the value range of AB is ()
There should be a process! The scope of problem solving knowledge should be below grade one (including grade one)
If the circle is paired with the line 2ax-by + 2 = 0, then the center (- 1,2) of the circle is on the line, i.e. a + B = 1, then y = a * b = a (1-A) = - A ^ 2 + A. according to the maximum value of the quadratic function, AB is obtained
Proof: no matter what real number x is, the value of algebraic formula x ^ 2-4x + 6 is always greater than zero
x²-4x+6
=x²-4x+4+2
=(x-2)²+2
∵(x-2)²≥0
∴(x-2)²+2＞0
The value of the algebraic formula X & # 178; - 4x + 6 is always greater than zero
Given x ^ 2-2y ^ 2 = 2, find the value of (2x-y) (x + 4Y) - 15
Xy = 0, forget to call
x²-2y²=2
(2x-y)(x+4y)-15=2x²+8xy-xy-4y²-15=2(x²-2y²)+7xy-15=7xy-11
xy=0
So = - 11
∵x^2-2Y^2=2 ∴x^2=2y^2+2
(2x-y)(x+4y)=2x(x+4y)-y(x+4y)=2x^+8xy-xy-4y^=2x^-4y^+7xy=2*2+7xy=4+7xy
4+7xy-15=7xy-11
As shown in the figure, given that the intersection point of graph of functions y = x + B and y = ax + 3 is p, then the solution set of inequality x + b > ax + 3 is ()
A. x＜1B. x＞1C. x≥1D. x≤1
The image intersection point of the functions y = x + B and y = ax + 3 is p, and the abscissa of P point is 1. According to the image, when x > 1, the image of the function y = x + B is on the top of the image of the function y = ax + 3, then the value of the function y = x + B is greater than the value of the function y = ax + 3, that is, the solution set x > 1 of the inequality x + b > ax + 3
3y-6 / 3Y + 5 = 2y-4 / 5y-4 to solve the equation
3y-6 / 3Y + 5 = 2y-4 / 5y-4
Multiply both sides of the equation by 6 (Y-2) to get
2(3y+5)=3(5y-4)
Then 6y + 10 = 15y-12 and y = 22 / 9
Test: substitute y = 22 / 9 into the original equation, left = right
So y = 22 / 9 is the solution of the original equation
Given x ^ 2-4x + y ^ 2-6y + 13 = 0, find x ^ 3Y ^ 2 + x ^ 4Y ^ 2
X ^ 2-4x + y ^ 2-6y + 13 = 0 can be changed into
(X-2)^2+(y-3)^2=0
So x = 2, y = 3
Then x ^ 3Y ^ 2 + x ^ 4Y ^ 2 = 8 * 9 + 16 * 9 = 216
X=2,Y=3
Then x ^ 3Y ^ 2 + x ^ 4Y ^ 2 = 8 * 9 + 16 * 9 = 216