What is the solution of the equation x squared minus 5x = 0? What is the solution of the equation x squared minus 9 = 0?

What is the solution of the equation x squared minus 5x = 0? What is the solution of the equation x squared minus 9 = 0?

The former is 0 or 5, the latter is ± 3
What is the solution of the equation x square = 5x?
x^2=5x
x^2-5x=0
x(x-5)=0
x1=0,x2=5
X^2-5X=0
X = 0 or x = 5
0 and 5
Primary school mathematics problem x + 3.5x = 9.9 the whole process of solving equation
x+3.5x=9.9
4.5x=9.9
x=9.9/4.5
x=2.2
4.5x=9.9
X = 2.02: where did 4.5 come from
Given that variables X and y satisfy the condition x ≥ 1 x − y ≤ 0 x + 2 y − 9 ≤ 0, the maximum value of 2x + y is ()
A. 3B. 6C. 9D. 12
Let x + y = 2x3 + y = 2x + 9 be the maximum number of feasible points in the graph
It is known that the quadratic function f (x) satisfies the conditions f (0) = 1 and f (x + 1) - f (x) = 2x. (1) find f (x); (2) find the maximum and minimum of F (x) in the interval [- 1, 1]
(1) Let f (x) = AX2 + BX + C, then f (x + 1) - f (x) = a (x + 1) 2 + B (x + 1) + C - (AX2 + BX + C) = 2aX + A + B ｜ from the problem C = 12ax + A + B = 2x ｜ 2A = 2A + B = 0C = 1 & nbsp; it is obtained that & nbsp; a = 1b = − 1C = 1 ｜ f (x) = x2-x + 1 (2) f (x) = x2-x + 1 = (x − 12) 2 + 34
The two sides of the rectangle are two root sign three + root sign two and two root sign three one root sign two respectively. Find the area and diagonal of the rectangle
The two sides of a rectangle are two root sign three + root sign two and two root sign three one root sign two respectively. Find the area and diagonal length of the rectangle
Rectangle, for rectangle, area = a * B (product of two sides) = (2 radical 3 + radical 2) * (2 radical 3-radical 2) = 10. Diagonal direct Pythagorean theorem: l (diagonal) = radical ((2 radical 3 + radical 2) * (2 radical 3 + radical 2) + (2 radical 3-radical 2) * (2 radical 3-radical 2)) = 2 radical 7
Let x + 2Y = 1, find the minimum value of x ^ 2 + y ^ 2; if x > 0, Y > 0, find the maximum value of x ^ 2 + y ^ 2
x=1-2y
∴x²+y²
=4y²-4y+1+y²
=5y²-4y+1
=5（y-0.4）²+0.2
So the minimum is 0.2
If x > 0, Y > 0
Then 1-2y > 0, Y > 0
∴0＜y＜0.5
When y infinitely approaches 0, the maximum value is 1
So it's close to infinity
x+2y=1
x=1-2y
x²+y²
=（1-2y）²+y²
=5（y-2/5）²+1/5≥1/5
That is, X & sup2; + Y & sup2; is the minimum of 1 / 5, x = 1 / 5, y = 2 / 5
X > 0, so 1-2y > 0, that is 0 < y < 1 / 2
For f (y) = 5 (Y-2 / 5) & sup2; + 1 / 5
f（0）=1
f（1/2）=1/4
The maximum value is close to 1
From x + 2Y = 1, y = 1 / 2 * (1-x)
By substituting the above formula into x ^ 2 + y ^ 2, we can get:
x^2+1/4*(1-x)^2=5/4*x^2-1/2*x+1/4
The minimum value of the above formula is obtained at x = 1 / 5
The minimum value is 5 / 4 * 1 / 25-1 / 2 * 1 / 5 + 1 / 4 = 1 / 5
If x > 0, Y > 0
Then 1-2y > 0, Y > 0
∴0＜y＜0.5
When y infinitely approaches 0, the maximum value is 1
So the maximum value is close to 1
Given that the quadratic function y = f (x) satisfies f (0) = 1 and f (x + 1) - f (x) = 2x, what are the maximum and minimum values of F (x) in the interval [- 1,1]
∵f(0)=1
Let f (x) = ax & # 178; + BX + 1
Then f (x + 1) = a (x + 1) &# 178; + B (x + 1) + 1 = ax & # 178; + (2a + b) x + A + B + 1
Then f (x + 1) - f (x) = 2aX + A + B = 2x
∴ 2a=2,a+b=0
∴ a=1,b=-1
f(x)=x²-x+1
f(x)=(x-1/2)²+3/4
x∈ 【-1,1】
When x = 1 / 2, y has a minimum value of 3 / 4
When x = - 1, y has a maximum value of 3
1 and 3
F1-f0 = 0, f0-f-1 = - 2 and so on
(root 5-1) to the power of 0 + (- 0.125) to the power of 2009 × 8 to the power of 2010
(root 5-1) to the power of 0 + (- 0.125) to the power of 2009 × 8 to the power of 2010
=1 + (- 0.125 × 8) to the power of 2009 × 8
=1-8
=-7
(root 5-1) to the power of 0 + (- 0.125) to the power of 2009 × 8 to the power of 2010
=1 + (- 1 / 8) to the power of 2009 × 8 to the power of 2010
=1-1 / 8 to the power of 2009 × 8 to the power of 2010
=1-8 to the power of - 2009 × 8 to the power of 2010
=1-8
= -7
Let m > 1, under the constraint y ≥ XY ≤ mxx + y ≤ 1, the maximum value of the objective function z = x + my is less than 2, then the value range of M is ()
A. （1，1+2）B. （1+2，+∞）C. （1，3）D. （3，+∞）
∵ m > 1, so the line y = MX intersects with the line x + y = 1 at (1m + 1, mm + 1), the line corresponding to the objective function z = x + my is perpendicular to the line y = MX, and at (1m + 1, mm + 1), the maximum value is obtained. The relationship is shown in the following figure: 1 + M2M + 1 < 2, the solution is 1-2 < m < 1 + 2, and ∵ m > 1, the solution is m ∈ (1, 1 + 2), so select: a
The constraint region is a triangle passing through the origin in the first quadrant, z = x + my = > x = - My + Z
Z can be regarded as the intercept of the line x = - My + Z on the x-axis
If you translate the line x = - my in the constraint area, it is not difficult to see that there is a minimum intercept at the origin, at this time z = 01
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