In the equation group ax + by = 16, BX + ay = 19, Xiao Ming misprints ① to x = 1, y = 7, Xiao Liang miscopies ② to x = - 2, y = 4, and gets the correct answer What is the solution of the original equations?

In the equation group ax + by = 16, BX + ay = 19, Xiao Ming misprints ① to x = 1, y = 7, Xiao Liang miscopies ② to x = - 2, y = 4, and gets the correct answer What is the solution of the original equations?

When solving the equations ax + by = 16 (1), BX + ay = 19 (2), Xiao Ming copied the equation (1) wrong and got the wrong solution x = 1, y = 7
X = 1, y = 7 are the correct solutions of BX + ay = 19
b+7a=19
Xiao Liang copied equation 2 wrong and got the wrong solution x = - 2, y = 4,
X = - 2, y = 4, is the correct solution of AX + by = 16
-2a+4b=16
b+7a=19 4b+28a=76
-2a+4b=16
30a=60
A=2
b=19-7a=19-7*2=5
So. A = 2, B = 5
2x+5y=16
5x+2y=19
X=3
Y=2
Known: ax + by = 3 ax * 2 + by * 2 = 7 ax * 3 + by * 3 = 16 ax * 4 + by * 4 = 42 ax * 5 + by * 5 =?
So troublesome, because AX2 + BY2 = 7, then AX2 = 7-by2, BY2 = 7-ax2, AX3 = 7x-bxy2, by3 = 7y-ax2y, AX3 + by3 = 7 (x + y) - XY (by + ax) = 16, namely 7 (x + y) - 3xy = 16, and because AX3 + by3 = 16, then AX3 = 16-by3, by3 = 16-ax3, ax4 = 16x-bxy3, by4 = 16y-ax3y, ax4 + by4 = 16 (x + y) - XY
High school mathematics y = 3x + 2 / X-1 range
The original function can be transformed into y = 5 / (x-1) + 3. It is equivalent to translating the image of function y = 5 / X one unit to the right and three units to the up. Because the value range of function y = 5 / X is R and not 0, the value range of original function is R and not equal to 3
Given that the solution of equation 3 (x-a) + 2 = x-a + 1 is suitable for inequality 2 (x-4) > 4a, the value range of a is obtained
If the solution of the equation is x = (2a-1) / 2 and satisfies the inequality, then 2 [(2a-1) / 2-4] > 4a
2a－1－8>4a
2A
The solution of the equation is: x = A-1 / 2
By introducing the inequality, we get 2 (A-1 / 2-4) = 2a-9
>4a
So: 2A < - 9
a< -9/2
If the solution of the equation is x = (2a-1) / 2, then 2 [(2a-1) / 2-4] > 4a
2a－1－8>4a
2a4a
a-9/2>2a
-9/2>a
The range of y = 3x / (x ^ 2 + 4)
I'm Gao Yisheng. Needless to say, it's too complicated,
When x = 0, y = 0
When X & gt; 0, y = 3x / (x * x + 4) = 3 / [x + (4 / x)]
From the basic inequality, when a & gt; 0, B & gt; 0, a + B & gt; = 2 √ [a * b], I will not prove it
y=3/[x+(4/x)]<=3/4
When X & lt; 0
y=-3/[-x+(4/-x)]>=-3/4
In conclusion, y belongs to [- 3 / 4,3 / 4]
Given that the integer solution of inequality system 3 (X-2) + 8 ＞ 2XX + 13 ≥ x-x-12 satisfies the equation AX + 6 = x-2a, find the value of A
To solve the system of inequalities: 3 (X-2) + 8 ＞ 2XX + 13 ≥ x-x-12, we get: - 2 ＜ x ≤ - 1, the integer solution of the system of equations is - 1, that is, x = - 1, which is substituted into the equation AX + 6 = x-2a to get - A + 6 = - 1-2a, and the solution is a = - 7
Find the range of y = 3x-1 in X ∈ [3,6],
Because y = 3x-1 is a monotone function
So when x = 3, Ymin = 3 * 3-1 = 8
When x = 6, ymax = 3 * 6-1 = 17
y∈[8,17]
If the solution set of inequality (2a-b) x > 3A + B about X is X
From the meaning of the title, 2a-b3a + B can be changed into:
x< (3a+b)/(2a-b)
Then: (3a + b) / (2a-b) = 7 / 3
9a+3b=14a-7b
5a=10b
The solution is a = 2B, B / a = 1 / 2
So: 4b-b
Y = 3x + 1 / x + 2 find the range of (1) x ≠ - 2 (2) x ≥ 0?
y=(3x+1)/(x+2)
=(3x+6-5)/(x+2)
=3-5/(x+2)
(1)
When x ≠ - 2, y ∈ R and Y ≠ 3
(2)
When x > = 0
x+2>=2
0=-5/2
3-5/(x+2)>=3-5/2=1/2
So Y > = 1 / 2
When the square of inequality X - 2aX + a > 0, X belongs to R constant, and the value range of a is obtained