Solve quadratic equation of one variable with matching method: 2x square + 1 = 3x

Solve quadratic equation of one variable with matching method: 2x square + 1 = 3x

2x^2-3x+1=0 (2x-1)(x-1)=0 x1=1/2 x2=1
If f (x) = = 2x + SiNx is a function defined on (- 1,1), then the solution set of inequality f (1-ah) + F (1-2a) less than 0 is. Only the final result~
Because f (x) = 2x + SiNx, f (x) is an odd function ∵ f (x) = 2 + cosx > 0, ∵ f (x) is an increasing function at (- 1,1) and f (1-A) + F (1-2a)
One root of the equation x2 + ax + B = 0 is 2, the other root is a positive number, and it is the root of the equation (x + 4) 2 = 3x + 52
The equation (x + 4) 2 = 3x + 52 is sorted out to be x2 + 5x-36 = 0, ∵ (x-4) (x + 9) = 0, ∵ X1 = 4, X2 = - 9, ∵ 4 is the root of the equation x2 + ax + B = 0. According to the relationship between the root and coefficient, we can get 2 + 4 = - A, 2 × 4 = B, ∵ a = - 6, B = 8
If FX = 2x + SiNx is a function defined on negative one to one, then the solution set of inequality f1-a + f1-2a less than 0 is
Can you get the picture
It is known that one solution of the quadratic equation x2 + MX + n = 0 is 2, the other solution is positive, and it is also the solution of the equation (x + 4) 2-52 = 3x. Can you find the values of M and N?
Solve the equation (x + 4) 2-52 = 3x, X2 + 8x + 16-52-3x = 0x2 + 5x-36 = 0, (x + 9) (x-4) = 0  X1 = - 9, X2 = 4, so the other root of the equation x2 + MX + n = 0 is 4. Substitute 2 and 4 into the equation x2 + MX + n = 0, and get: 4 + 2m + n = 0 & nbsp; & nbsp; & nbsp; ① 16 + 4m + n = 0 & nbsp; & nbsp; ② the solution is
Given the function f (x) = 2, x > 1; (x-1) &# 178; + 1, X ≤ 1, then the solution set of the inequality f (1-x & # 178;) > F (2x) is?
Need detailed process, thank you
First, judge the value range of G (x) = 1-x ^ 2, it is easy to know that when the value range is y = x > = 0, 3 / 40
(x^2-2x+1)(x^2+2x-1)>0
x^2+2x-1>0
x> When - 1 + √ 2 (x = x > = 0), the solution set of the original inequality is - 1 + √ 21
x> When x 1 / 2, the solution set of the original inequality is x > 1
(3) When X0
(x^2-2x+1)(x^2+2x-1)>0
x^2+2x-1>0
X-1 + √ 2 (rounding off)
So, when x
Because x ≤ 1,; f (x) = (x-1) &# 178; + 1
So 1-x & # 178; 1, f (2x) = 2
(x-1)²+1>2
(x-1)²>1
x1>2,x2
If the square of MX + 3x-4 = the square of 3x is a quadratic equation of one variable, then the value range of M is?
There are some missing words in the title
The main topic is:
The square of MX + 3x-4 = the square of 3x is a quadratic equation of one variable about X,
Sort it out
(m-3)x^2+3x-4=0
The quadratic equation of one variable can be solved as long as the coefficient of quadratic term is not 0
m≠3
We don't need to consider too much here... The problem doesn't ask the solution of the equation. We don't need to consider the discriminant of the root at all
(M-3) x + 3x-4 = 0
Because it's a quadratic equation of one variable, so
m-3≠0
Namely
m≠3
Since it is a quadratic equation of one variable. The coefficient of quadratic term cannot be 0
So m ≠ 3
The square of MX + 3x-4 = the square of 3x
Square of (M-3) x + 3x-4 = 0
∴m-3≠0
That is, m ≠ 3
m-3≠0，
m≠3
That's it.
The function f (x) = SiNx + sin (x + π / 2) x is a real number
If f (a) = 3 / 4, find the value of sin2a
f(X)=sinx+sin(X+π/2）
=sinx+cosx
f(a)=sina+cosa
So Sina + cosa = 3 / 4
sin²a+2sinacosa+cos²a=9/16
1+sin2a=9/16
So sin2a = 9 / 16-1 = - 7 / 16
-7/16
When k takes what value, the quadratic equation 2x2-3x-k = 0 has two positive real roots
Namely
Δ=3^2+8k>0
x1x2>0→ -k/2>0
in summary,
-9/8
－9／8＜k＜0
Given the function f (x) = 2x + SiNx, X belongs to R, and f (1-A) + F (2a)
Since f (x) = 2x + SiNx and f (- x) = - 2x + sin (- x) = - f (x), f (x) is an odd function
∵ F & # 180; (x) = 2 + cosx > 0, ∵ f (x) is an increasing function on R
F (1-A) + F (2a)
If x0 belongs to [1 / E, e], f (x0) & gt; = g (x0), then f (x) - GA & gt; = 2E, the function increases in [1 / E, 1] and decreases in [1, e], so max = H (1) & gt; = 0, the solution is a & lt;