4(x-1)²-9(3-2x)²=0

4(x-1)²-9(3-2x)²=0

4(x-1)²-9(3-2x)²=0
[2 (x-1)] square - [3 (3-2x)] square = 0
(2(x-1)+3(3-2x))(2(x-1)-3(3-2x))=0
(-4x+7)(8x-11)=0
X = 7 / 8 or x = 7 / 8
4(x-1)²-9(3-2x)²=0
(2x-2-9+6x)(2x-2+9-6x)=0
(8x-11)(7-4x)=0
X = 11 / 8 or 7 / 4
The solution equation (1) 25 (2x-5) & # 178; - 9 (x + 4) & # 178; = 0 (2) 2x (x-3) = 5 (x-3)
(3) (X-5) (x + 2) = 8, the whole process must be urgent
（1） 25（2x－5）²－9﹙x＋4﹚²＝0
[5(2x-5)+3(x+4)][5(2x-5)-3(x+4)]=0
(13x-13)(7x-37)=0
∴x=1 x=37/7
﹙2﹚ 2x﹙x－3﹚＝5﹙x－3﹚
(x-3)(2x-5)=0
x=3 x=5/2
﹙3﹚ ﹙x－5﹚﹙x＋2﹚＝8
x²-3x-18=0
(x-6)(x+3)=0
∴x=6 x=-3
Give additional reward points and ask me, 5 minutes to solve the problem
9（x-3）（x+4）=（2x-1）（2x+1）+5（x-2）2．
The equation is arranged as follows: 9x2 + 9x-108 = 4x2-1 + 5x2-20x + 20, the combination of terms is 29x = 127, and the solution is x = 12729
Given the square of the equation x about X - (M + 2) x + (2m-1) = 0, it is proved that the equation always has two unequal real roots
(1) It is proved that: ∵ (?) = (M + 2) 2-4 (2m-1) = (m-2) 2 + 4, ∵ (m-2) 2 + 4 ≥ 4, that is, ∵ = (M + 2) 2-4 (2m-1) = 0, has two unequal real roots for the equation X2 - (M + 2) x + (2m-1) = 0 of X. according to the meaning of the question, we get 12-1 × (M + 2) + (2m-1) = 0, and the solution is
A solution equation. (x / 4) ^ 2 + (56-x / 4) ^ 2 = 100
Let y = x / 4;
The original formula is changed to y ^ 2 + (56-y) ^ 2 = 100
Expand 2Y ^ 2-2 * 56Y + 3036 = 0
y^2-56y+1518=0
(y-23)(y-33)=0
y1=23;y2=33
Substituting y = x / 4, we get
x1=92;x=132
-759/7
X*X/8+3136-28X=100
X*X-28X+24288=0
unsolvable
It is known that the equation x ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 about X has two equal real roots. Try to judge whether the straight line y = (2m-3) x-4m + 7 passes a (- 2,4) and explain the reason
∵ x ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two unequal real roots
That is, (2m + 1) ^ 2-4 (m ^ 2 + 2) > 0
The solution is m > 7 / 4
∴ 2m-3>0 ,-4m+7
On the unary linear equation of X, the square of x plus 2 (k minus 1) the square of x plus K minus 1 is equal to zero. There are two unequal roots of real numbers. Is it possible that 0 is a root of the equation?
There are two unequal real roots
The discriminant is greater than 0
4(k-1)²-4(k²-1)>0
k²-2k+1-k²+1>0
K
3（x+y)=（8/x+2/y)(x+y)
=10+(8y/x)+2x/y)
x/y>0,y/x>0
So 8y / x) + 2x / Y ≥ 2 √ (8y / x) * 2x / y) = 8
So 3 (x + y) ≥ 10 + 8 = 18
x+y≥6
The minimum is 6
56 △ x = 24 solution equation
56 △ x = 24 solution equation
x=56÷24
x=7/3
56 = 24x x = 56 / 24 answer 3 / 7
56÷x=24
56=24x
x=56/24
x=7/3
56÷X＝24
56=24x
x=56÷24
x=7/3
56÷x=24
x=56÷24
X = 7 / 3
It is known that the equation x2 + (2m + 1) x + M2 + 2 = 0 about X has two unequal real roots, which is to judge whether the straight line y = (2m-3) x-4m + 7 can pass through a (- 2,4), and say
And explain the reason
There are two unequal real roots, and the result is m > 7 / 4. How can we calculate them later?
Substitute x = - 2, y = 4
y=（2m-3)x-4m+7
4=-2(2m-3)-4m+7
-4m+6-4m+7=4
8m=9
m=9/8
But M > 7 / 4
dissatisfaction
therefore
Straight line y = (2m-3) x-4m + 7 cannot pass through a (- 2,4)
Substitute x = - 2, y = 4
y=（2m-3)x-4m+7
4=-2(2m-3)-4m+7
-4m+6-4m+7=4
8m=9
m=9/8
But M > 7 / 4
dissatisfaction
therefore
Straight line y = (2m-3) x-4m + 7 cannot pass through a (- 2,4)
When k takes what value, the square of the equation KX - x + 1 = 0 has no real root process
Obviously, when k = 0, the equation has real roots;
When k is not equal to 0,
b^2-4ac=1-4k1/4
From the meaning of the title, (- 1) &# - 4K < 0 and K ≠ 0
The solution is k > 1 / 4