In mathematics of grade two of junior high school, X & # 178; - x + 1 = O, X & # 178; - 1ox + 25 = 0, do not understand the equation and judge the following equation

In mathematics of grade two of junior high school, X & # 178; - x + 1 = O, X & # 178; - 1ox + 25 = 0, do not understand the equation and judge the following equation

1)△=b²-4ac=1-4*1*1=-3＜0
So this equation has no solution
2)△=b²-4ac=100-4*1*25=0
So this equation has two equal real roots
Calculate (x + y) & # 178; (X-Y) & # 178; + (2x-y) & # 178; (2x + y) & # 178;
﹙x＋y﹚²﹙x－y﹚²＋﹙2x－y﹚²﹙2x＋y﹚²=(x²-y²)²+(4x²-y²)²=x⁴-2x²y²+y⁴+16x⁴-8x²y²+y⁴=17x⁴-10x²...
17x^4+2y^4-10 X^2 Y^2
His answer is good downstairs.
Calculate [2x & # 178; - (X-Y) (x + y)] [(2-x) (2 + x) + (Y-2) (y + 2)]
[2x²-(x-y)(x+y)][(2-x)(2+x)+(y-2)(y+2)]
=[2x²-x²+y²][4-x²+y²-4]
=[x²+y²][y²-x²]
=y^4-x^4
=（2x^2-x^2+y^2)(4-x^2+y^2-4)=(x^2+y^2)(y^2-x^2)=y^4-x^4
Original formula = [2x * x - (x * X-Y * y)] [4-x * x + y * y-4]
=（y*y-x*x）（y*y-x*x）
=（y+x）（y+x）（y-x）（y-x）
The second power of (X-2) - 25 = 0 (direct flattening method)
（X-2)²-25=0
（X-2)²=25
x-2=±5
So x = - 3 or x = 7
If you don't understand, I wish you a happy study!
The quadratic equation of one variable x 2-6x-5 = 0 is reduced to the form of (x + a) 2 = B
Solving the original equation can be reduced to x2-6x = 5, formula x2-6x + 9 = 5 + 9, that is (x-3) 2 = 14
Through point P [2,1], make the square of circle x plus the square of y-2x-2y plus the tangent of 1 = 0, and try to find the tangent equation
The equation of circle is as follows:
(x-1)^2+（y-1)^2=1
In the plane rectangular coordinate system, draw the circle image and point P
Since the center of the circle is (1,1)
Same as the ordinate of point P
Therefore:
One tangent is:
X=2
And because P is on the circle,
There is only one tangent
Therefore:
Tangent equation:
X=2
Some people say that no matter what real numbers x and Y take, the value of the algebraic formula X & sup2; + Y & sup2; - 10x + 8y + 45 is always positive. What do you think,
x²+y²-10x+8y+45
=x²-10x+25+y²+8y+16+4
=(x-5)²+(y+4)²+4
Because (X-5) & sup2; ≥ 0, (y + 4) & sup2; ≥ 0
So (X-5) & sup2; + (y + 4) & sup2; + 4 ≥ 4
Algebraic formula X & sup2; + Y & sup2; - 10x + 8y + 45 ≥ 4
That is, only if x = 5 and y = - 4, X & sup2; + Y & sup2; - 10x + 8y + 45 = 4,
Under other conditions, the value is more than 4
x²+y²-10x+8y+45
=x²-10x+25+y²+8y+16+4
=(x-5)²+(y+4)²+4
Because (X-5) & sup2; ≥ 0, (y + 4) & sup2; ≥ 0
So (X-5) & sup2; + (y + 4) & sup2; + 4 ≥ 4
therefore
The value of the algebraic expression X & sup2; + Y & sup2; - 10x + 8y + 45 is always positive
You can find the formula:
(X-5) square + (y + 4) square + 4 is greater than or equal to 1
Given that real numbers x and y satisfy x2 + xy-y2 = 0, find the value of X and y
x^2+xy-y^2=0
x(x+y)=y^2
(x+y)/y=y/x
(x/y)+1=y/x
Let X / y = M
m+1=1/m
m^2+m-1=0
Solution M
You divide the equation by the square of Y Wait until the quadratic equation of X / Y! Then we can get X / y
2 (x + 1) square = 24 + flattening method
X1 = double root 3-1 x2 = negative double root 3-1
If the quadratic equation x2-6x-5 = 0 is changed into the form of (x + a) 2 = B, then B = ()
A. -4B. 4C. -14D. 14
∵ x2-6x-5 = 0, ∵ x2-6x = 5, ∵ x2-6x + 9 = 5 + 9, ∵ x-3) 2 = 14. ∵ B = 14