(x-1) ^ 2 + 3, (X-2) ^ 2 + 2x, (1 / 2 X-2) ^ 2 + 3 / 4 x ^ 2 is three different forms of formula of x ^ 2-2x + 4. Find three different forms of formula of x ^ 2-4x + 2

(x-1) ^ 2 + 3, (X-2) ^ 2 + 2x, (1 / 2 X-2) ^ 2 + 3 / 4 x ^ 2 is three different forms of formula of x ^ 2-2x + 4. Find three different forms of formula of x ^ 2-4x + 2

(x-2)^2-4
(2x + root 2) ^ 2 - (2 root 2 + 4) x
(root 2x-root 2) ^ 2-x ^ 2
x²-4x+2
=x²-4x+4-2
=(x-2)²-1
x²-4x+2
=x²+2√2x+2-4x-2√2x
=(x+√2)²-(4-2√2)x
x²-4x+2
=2x²-4x+2-x²
=2(x-1)²-x²
Can you make formula 2x ^ 2 + 4x-3 into a (x + k) ^ 2 + B
2x^2+4x-3
=2(x^2+2x)-3
=2(x^2+2*1*x+1^2-1^2)-3
=2(x+1)^2-5
x^2+ax+b
=[x^2+2*a/2*x+(a/2)^2-(a/2)^2]+b
=(x+a/2)^2+b-a^2/4
2(x+1)^2-5
2x^2+4x-3
=2(x^2+2x)-3
=2(x^2+2*1*x+1^2-1^2)-3
=2(x+1)^2-5
x^2+ax+b
=[x^2+2*a/2*x+(a/2)^2-(a/2)^2]+b
=(x+a/2)^2+b-a^2/4
It's simple
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Factorization 3x ^ 2Y ^ 2-11xy + 10
And (2) 2x ^ 2-7xy-22y ^ 2
(3) (x^2+2x)^2-7(x^2+2x)-8
(4) 1999x ^ 2 - (1999 ^ 2-1) x-1999 please help,
(1)3x^2y^2-11xy+10=(xy-2)(3xy-5)
(2) 2x^2-7xy-22y^2 =(x+2)(2x-11)
(3) (x^2+2x)^2-7(x^2+2x)-8=(x^2+2x-8)(x^2+2x+1)=(x+4)(x-2)(x+1)(x=1)
(4) 1999x^2-(1999^2-1)x-1999=(1999x-1)(x+1999)
3x^2y^2-11xy+10 =(3xy+a)(xy+b)=3x^2y^2+(a+3b)xy+ab
So AB = 10
a+3b=-11
The solution is a = - 5, B = - 2
So 3x ^ 2Y ^ 2-11xy + 10 = (3x-5) (X-2)
The rest of the problems can be solved in the same way
（2）2x^2-7xy-22y^2 =（2x-11y）（x+2y)
(3) ^ 2 + 2x... Expand
3x^2y^2-11xy+10 =(3xy+a)(xy+b)=3x^2y^2+(a+3b)xy+ab
So AB = 10
a+3b=-11
The solution is a = - 5, B = - 2
So 3x ^ 2Y ^ 2-11xy + 10 = (3x-5) (X-2)
The rest of the problems can be solved in the same way
（2）2x^2-7xy-22y^2 =（2x-11y）（x+2y)
(3) (x^2+2x)^2-7(x^2+2x)-8 =(x^2+2x-8）（x^2+2x+1)=(x+1)^2(x^2+2x-8)
(4) 1999x ^ 2 - (1999 ^ 2-1) x-1999 = (1999x + 1) (x-1999) put away
The solution set of known inequality system x-3ax-4 is 4
A=7
First calculate 3 (x-4) > x-4. Expand the formula on the left and get 3x-12 > x-4. Continue to simplify and get 2x > 8, so x > 4
Then according to the known solution set 4
x-3ax-4
2x>8
X>4
So 4
Factorization X & sup2; - 5xy + 6y & sup2; = () ()
x²-5xy+6y²=(x-2y)(x-3y)
x²-5xy+6y²=(x－2y)(x－3y)
If the system of inequalities about X: x > A + 2, X
a+24
A>2
Y = - x + a when x2
So the image of the function y = - x + A does not pass through the third quadrant
If the system of inequalities about X: x > A + 2, X
[(3x + 4Y) ^ 2 - 3x (3x + 4Y)] / (- 6y) where x = 1, y = - 3, simplification and evaluation
[(3x+4y)^2 - 3x(3x+4y)] / (- 6y) =[(3x+4y)(3x+4y-3x)]/(-6y) =(3x+4y)*4y/(-6y) =-2/3*(3x+4y) =-2/3*(3-12) =6
In the system of linear inequalities with one variable of mathematical inequalities and systems of inequalities, how to solve the application problems of solving linear inequalities with one variable? I always make mistakes in symbols
The linear inequality with one variable is basically the same as the linear equation with one variable
If both sides of the inequality multiply or divide by a negative number at the same time, the sign will be reversed
When ＞ or ≥ is changed into ＜ or ≤, when ＜ or ＜ is changed into ＞ or ≥
[(3x + 4Y) - 3x (3x + 4Y)] / (- 6y), where x = - 1, y = 3
Degenerate evaluation
[﹙3x+4y﹚-3x﹙3x+4y =(12xy+16y)÷﹙-6y﹚ =-2x-8/3y =2-8 =-6
How to find the range of value for solving linear inequality of one variable under 8?
Just like the linear equation with one variable, you need to find the result of the unknown. But if both sides of the inequality divide or multiply by a negative number, the direction of the inequality will change
I'm sorry, I won't either, because I'm in grade one now