It is known that the solution of equation 5x = 2x-2a is inequality 3 (x + 2) - 7

It is known that the solution of equation 5x = 2x-2a is inequality 3 (x + 2) - 7

3(x+2)-7

2X ^ 4-3x ^ 3 + ax ^ 2 + 5x + B can be divided by X + 1 and X-2 to find a and B

Let f (x) = 2x ^ 4-3x ^ 3 + ax ^ 2 + 5x + B = K (x + 1) (X-2)
That is, when x = - 1 and x = 2, the polynomial value f (x) is 0
Substitute x to get
2+3+a-5+b=0 a+b=0
32-24+4a+10+b=0 4a+b=-18
The solutions are a, B

The difference between a number and its reciprocal is 14 / 15?

15. Let this number be x; X-1 / x = (x ^ 2-1) / x = (x + 1) (x-1) / X
14 of 15 = (14 * 15 + 14) / 15 = (14 + 16) / 15 = (15 + 1) (15-1) / 15
It is deduced that the number x = 15

F (x1, X2, x3) = x1x2 + x1x3 + x2x3 find the standard form by matching method

There is no square term in quadratic form, so the square term should be constructed by a coordinate transformation
Let X1 = Y1 + Y2
x2=y1-y2
x3=y3
f=(y1+y2)(y1-y2)+(y1+y2)y3+(y1-y2)y3
=y1²-y2²+2y1y3
=y1²+2y1y3+y3²-y2²-y3²
=(y1-y3)²-y2²-y3²
The standard form is: F = Z1 & # 178; - Z2 & # 178; - Z3 & # 178;

Given the function f (x) = - x2 + 2ex + M-1, G (x) = x + E2x & nbsp; (x > 0); (1) if G (x) = m has real roots, find the value range of M; (2) determine the value range of M, so that G (x) - f (x) = 0 has two different real roots

(1) Method 1: ∵ g (x) = x + E2x ≥ 2E, the condition of equal sign is x = E. so the value range of G (x) is [2E, + ∞), so if M ≥ 2E, then G (x) = m has real root. So the value range of M is {m | m ≥ 2E}. Method 2: make the image of G (x) = x + E2x & nbsp; (x > 0) as shown in the figure

It is known that M is the root of the quadratic equation of one variable X & # 178; - 2x-1 = 0. Solve the equation and find the value of 2m & # 178; - 4m + 1

Because m is the root of the quadratic equation x & # 178; - 2x-1 = 0
So m ^ 2-2m-1 = 0
So m ^ 2-2m = 1
So 2m & # 178; - 4m + 1
=2（2m^2-2m）+1=2*1+1=3

What is the symmetry axis of quadratic function y = 0.5x square?

Y-axis
If it's y = 0.5x square + 1, it's a straight line x = 1

4X minus 5 times 12 equals 6
No brackets

4X-5×12=6
4X-60=6
4X=6+60=66
X=66÷4
X=16.5

The quadratic equation of one variable x ^ 2 - (K + 1) x + 1 / 4K ^ 2 + 1 = 0 is known
1. What is the value of K, the equation has two real roots
2. If two real roots X1 and X2 of the equation satisfy that the absolute value of X1 = X2, find the value of K

1,△=K^2+2k+1-k^2-4=2k-3≥0
∴k≥3/2
2, the absolute value of X1 = X2, then x2 ± X1 = 0
∵x1+x2=k+1,x1x2=1/4k^2+1
∴(x1-x2)^2=(x1+x2)^2-4x1x2=2k-3
∴x1-x2=±√2k-3
When X1 + x2 = 0, K + 1 = 0, k = - 1 (rounding)
When x2-x1 = 0, ± √ 2k-3 = 0, k = 3 / 2
When k = 3 / 2, two real roots X1 and X2 satisfy the absolute value of X1 = x2

9.6-x = 5.3 + X 49-4 (x-1) = 70.9 + x = 0.6 (2.5-x)
9.6-x = 5.3 + X 49-4 (x-1) = 70.9 + x = 0.6 (2.5-x)

9.6-x=5.3+x
2x=9.6-5.3
2x=4.3
x=2.15
49－4(x-1)=7
49－4x+4=7
4x=49+4-7
4x=46
x=11.5
0.9+x=0.6(2.5-x)
0.9+x=1.5-0.6x
1.6x=1.5-0.9
1.6x=0.6
x=0.375