(x×x+3x+2)(x×x+7x+12)-24

(x×x+3x+2)(x×x+7x+12)-24

(x×x+3x+2)(x×x+7x+12)-24
=(x+1)(x+2)(x+3)(x+4)-24
=(x+1)(x+4)(x+2)(x+3)-24
=(x²+5x+4)(x²+5x+6)-24
=(x²+5x)²+10(x²+5x)+24-24
=(x²+5x)²+10(x²+5x)
=(x²+5x)(x²+5x+10)
=x(x+5)(x²+5x+10)

3x + 16 = x + 40, find X

3x-x=40-16
2x=24
x=12

x+(3x-16)/25%=40

x+(3x-16)/25%=40
25% on both sides
0.25x+3x-16=10
0.25x+3x=10+16
3.25x=26
x=26÷3.25
x=8

We know that the quadratic equation 2x + 3y-4 = 0, where x and y are opposite to each other

X and y are opposite numbers = = > y = - X
Bring in have
2x-3x-4=0
x=-4
y=4

Given the quadratic equation 2x-3y = 5, the solution of this equation is obtained as follows. (1) the value of X is 1 times larger than that of Y

2x-3y=5 ①
x=3y+1 ②
Substitute (2) for (1)
2×（3y+1）-3y=5
6y+2-3y=5
3y=3
y=1
Substituting y = 1 into 2
x=4
therefore
x=4
y=1

When k takes what value, the equation 3 (xsquare) - 2 (3K + 1) x + 3 (ksquare) - 1 = 0 1 has one following zero; 2 has two equal real roots

(1) Because x = 0
So 3 * 0-2 * (3K + 1) * 0 + 3K ^ 2-1 = 0
3k^2-1=0
K = radical 3 / 3 or K = - radical 3 / 3
(2) Because there are two equal real roots
So [- 2 (3K + 1)] ^ 2-4 * 3 * (3K ^ 2-1) = 0
36k^2+24k+4-36k^2+12=0
24k=-16
k=-2/3

If the equation x ^ 2 + 2 (k-1) x + 3K ^ 2-11 = 0 has no real root, the value range of the real number k will be zero

Because there is no real number root real number
So △ 0
That is B & sup2; - 4ac < 0
[2（k-1）]²-4（3k²-11）＜0
-8k²-8k＋48＜0
8k²＋8k－48＞0
k²＋k－6＞0
The solution is k ＜ - 3 or K ＞ 2

The equation X-2 (x + 2x + a (X-2) + X | X-2 = 0 has only one real root
That bar is a division sign, a fraction line. Because there's no other kind of score

（x-2)/x+(2x+a)/[x(x-2)]+x/(x-2)=0
(x-2)^2+2x+a+x^2=0
2x^2-2x+4+a=0
When a = - 4, 2x ^ 2-2x = 0,
X = 0 or x = 2, x = 0 is the increasing root,
So when a = - 4, x = 0
When a ≠ - 4, Δ = 4-8 (a + 4) = 0
a=-7/2,x=1/2
So when a is - 4 or - 7 / 2, the equation has only one real root

When a is a value, the equation x / (X-2) - (X-2) / X - (2x + a) / X (2-x) = 0 has only one real root?

First, we get (6x-4 + a) / X (X-2) = 0, where x cannot be 0 and 2, then 6x-4 + a = 0, when x = 0, a = 4, when x is 2, a = - 8, so when a is not equal to 4 and - 8, X has only one real root (4-A) / 6

If a > 2, then how many roots of the equation 1 / 3 x cube - ax square + 1 = 0 on (0,2)?

Let f (x) = 1 / 3x ^ 3-ax ^ 2 + 1
f(0)=1
f'(x)=x^2-2ax=x(x-2a)
On (0,2), because a > 2, f '(x) 0
According to the intermediate value theorem, f (x) has no real root at (0,2)