X-3 / 5x = 6 and 2 / 5

X-3 / 5x = 6 and 2 / 5

X-3 / 5x = 6 and 2 / 5
2/5x=32/5
x=32/5÷2/5
x=16
Sixteen
Original formula: 2 / 5x = 32 / 5
X=16
5x^2-32x-3=0
A set of root formula will come out
How to solve the equation of 3 / 5x + 6 = X-6
3/5x+6=x-6
x-3/5x=6+6
2/5x=12
x=30
3x/5=x-12
Multiply both sides by 5 at the same time
3x=5x-60
2x=60
I don't understand why
Solve the equation x + 5 / 3 + 4 = x + 3 / 2-5x-2 / 6
5X=3/2-2/6-5/3-4,5X=3/2-1/3-5/3-4,5X=3/2-2-4,5X=3/2-6,5X=-9/2,X=-0.9
When a is a, the solution of equation 4 (x + 2) - 5 = 3A + 2 is equal to the solution of equation (3a + 1 / 3) x = a (2x + 3) / 2?
The solution of equation 4 (x + 2) -- 5 = 3A + 2 is: x = (3a -- 1) / 4,
The solution of the equation (3a + 1 / 3) x = a (2x + 3) / 2 is: x = 9A / (12a + 2),
Because the solutions of the above two equations are equal,
So (3a -- 1) / 4 = 9A / (12a + 2)
(3a--1)(12a+2)=36a
36a^2--6a--2=36a
36a^2--42a--2=0
A1 = (7 + radical 57) / 12, A2 = (7 -- radical 57) / 12
X-Y of X + 2Y △ x2 + 4xy + x2-y2 of 4y2
X-Y of X + 2Y △ x2 + 4xy + x2-y2 of 4y2
=(x-y)/(x+2y)* (x+2y)²/[(x+y)(x-y)]
=(x+2y)/(x+y)
Given the quadratic function y = x & # 178; + 2x + m, if x = - 1 has the minimum value - 2, then M=
-1
Y = x & # 178; + 2x + M = (x + 1) &# 178; + M-1, when x + 1 = 0, the function has the minimum value, the minimum value = M-1, according to the meaning of the problem, M-1 = - 2, the solution is m = - 1.
If LAL = 5 and LBL = 2, what is a + B
LAL = 5, LBL = 2, so a = 5 or - 5, B = 2 or - 2
Then a + B = 7 or 3 or - 3 or - 7
lal=5,lbl=2
a=±5
b=±2
A + B = 7 or 3 or - 3 or - 7
a+b=7，-7,3，-3
lal=5,lbl=2
Then a = ± 5, B = ± 2
a+b=5+2=7
a+b= -5+2= -3
a+b=5+（-2）=3
a+b= -5+（-2）= -7
If the solution of the equation two thirds (x-a-3) = three thirds (2x-3a + 2) is x = A-1, find the value of a (change x to A-1)
I'm a = 6, right!
A=6
Given the function f (x) = - x2 + 4x + A, X ∈ [0, 1], if f (x) has a minimum value of - 2, then the maximum value of F (x) is ()
A. 1B. 0C. -1D. 2
The function f (x) = - x2 + 4x + a = - (X-2) 2 + A + 4 ∵ x ∈ [0, 1], the function f (x) = - x2 + 4x + a monotonically increases on [0, 1]. When x = 0, f (x) has the minimum value f (0) = a = - 2. When x = 1, f (x) has the maximum value f (1) = 3 + a = 3-2 = 1, so we choose a
The quadratic function y = - X & # 178; + 2x + 3 is known
1. When x ∈ R, find the range of F (x)
2. When x ∈ [- 1,0], find the range of F (x)
3. When x ∈ [0,3], find the range of F (x)
4. When x ∈ [a, a + 2], find the maximum of F (x)
5. When x ∈ [a, a + 2], find the minimum value of F (x)
1.4 to negative infinity
2.0 to negative infinity
3.0 to 4.0
4. If a is greater than 1, then - A ^ 2-2a + 3 to - A ^ 2 + 2A + 3
If a is less than - 1, then - A ^ 2 + 2A + 3 to - A ^ 2-2a + 3
If 1 ＞ a ＞ - 1
The maximum is 4
If a is less than 0, the minimum is - A ^ 2 + 2A + 3
If a is greater than 0, the minimum is - A ^ 2-2a + 3