It is proved that the equation x & sup2; + (k-1) x + (K-3) = 0 with respect to X has two unequal real roots

It is proved that the equation x & sup2; + (k-1) x + (K-3) = 0 with respect to X has two unequal real roots

△＞0
（k-1）^2-4(k-3)＞0
k^2-2k+1-4k+12＞0
k^2-6k+13＞0
k^2-6k+9+4＞0
（k-3）^2+4＞0
Because (K-3) ^ 2 ≥ 0
So △ = (K-3) ^ 2 + 4 > 0

It is proved that the equation (x-a) (X-B) = 1 has two unequal real roots

Expand, x ^ 2 - (a + b) x + AB-1 = 0
Discriminant = (a + b) ^ 2-4ab + 4
=(a-b)^2+4>0
There are two unequal roots

It is proved that the quadratic equation (x + 1) (X-2) = K2 with respect to X has two unequal real roots

The original equation is: (x + 1) (X-2) = K2
x²－x－2=k²
x²－x－(2＋k²)=0
The discriminant ⊿ = 1 + 4 (2 + K & # 178;)
∵k²≧0
∴1＋4(2＋k²)＞0
So the original equation has two unequal real roots

It is proved that no matter what the value of K is, the equation (x + 1) (x-3) = k2-3 has two unequal real roots

（x+1）（x-3）=k²-3
x²－3x＋x－3－k²＋3=0
x²－2x－k²＝0
⊿＝﹙－2﹚²－4×1×﹙－k²﹚＝4＋4k²≧4﹥0
Whatever the value of K, the equation (x + 1) (x-3) = k2-3 has two unequal real roots

As shown in the figure, ab ∥ CD, BF bisection ﹥ Abe, DF bisection ﹥ CDE, ﹥ BFD = 55 ° to calculate the degree of ﹥ bed

As shown in the figure, the over point E and F are respectively made as eg 8741; AB, FH \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ ∠ 1 + 2 ∠ 2 = 2 (∠ 1 + 2) = 2 × 55 ° = 110 °

Given three points a (1,1) B (- 1,0) C (0,1), find another point d (x, y) so that vector AB = vector CD

It's easy to know,
Vector AB = (- 2, - 1)
Vector CD = (x, Y-1)
∴（-2,-1）=(x,y-1)
-2=x,-1=y-1.
∴x=-2,y=0
∴D(-2,0)

In RT △ ABC, ∠ a = 45 ° and ∠ C = 90 °, point D is on segment AC, ∠ BDC = 60 ° ad = 1, BD = ()

Ah. Landlord. Khan. Isn't that the median line of the right angle side? Isn't it equal to the general of the hypotenuse side?

Let the two focuses of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) be F1F2. If there is a point m on the ellipse such that f1pf2 = 120 °, try to find the eccentricity of the ellipse
Let the two focuses of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) be F1 and F2. If there is a point m on the ellipse such that the angle f1pf2 = 120 °, try to find the eccentricity of the ellipse?
Range of eccentricity

∵ when m is on y-axis, ∠ f1mf2 is maximum
When m is on y-axis, ∠ f1mf2 ≥ 120 °, then ∠ oaf1 ≥ 60 °
Sin ∠ oaf1 ≥ sin60 ° = (root 3) / 2
Then E = C / a = sin ∠ oaf1 ≥ sin60 ° = (radical 3) / 2
The eccentricity of ellipse is less than 1
(radical 3) / 2 ≤ e ≤ 1

In the triangle ABC, ab = AC = 3, BC = 2, find the value of sina, SINB

Ad ⊥ BC in D, CE ⊥ AB in E
∵AB=AC
AD⊥BC
Ψ BD = CD = 1 (three wires in one)
The Pythagorean theorem ad = √ (3 & # 178; - 1 & # 178;) = 2 √ 2
sinB=AD/AB=2√2/3
Δ ABC area = 1 / 2 * ad * BC = 1 / 2 * ce * ab
∴CE=4√2/3
∴sinA=CE/AC=4√2/9

The minimum value of function y = (X & # 178; - 2x) & # 189; + 2 Λ (X & # 178; - 5x + 4) & # 189;)

The definition field is: X "0, or X" 4
When x = 4, the minimum is: 1 + 2 root sign 2