1. It is known that for K ∈ R, the line y = KX + 1 and the ellipse (x ^ 2) / 5 + (y ^ 2) / M = 1 always have a common point, then the value range of the real number m is_______ . How can I figure it out wrong A (0,1) B (0,5) C [1,5] ∪ (5, positive infinity) d [1,5] I figured it out. It's wrong for me

Choose C
If the line passes (0,1), if the point is in the ellipse, then there is always a common point, so m > = 1. Because it is an ellipse, M is not equal to 5, choose C
Known: 10 + square root 3 = x + y, where x is an integer, and O < y < 1, find the opposite number of X-Y
One more
In the plane rectangular coordinate system, a coordinate is (root 3-root 2,0), C coordinate is (- root 3-root 2,0), B is on the Y axis, and s of triangle ABC (I forgot whether it is area or perimeter) = root 3. Move ABC along the X axis, root 2 units long, and point A.B.C moves to a ', B', C '
1. Find the coordinates of B
2. Find the coordinates of a '. B'. C '
3. Find the area of quadrilateral c'abb '
emergency
X = 11, y = square root 3 - 1, X - y = 11 - (square root 3 - 1) = 12 - square root 3
Find out the length of AC: root 3 - root 2 - (- root 3 - root 2) = 2 root 3. Then you think about the perimeter or area, find out the coordinates of point B, and draw a picture
If the line y = KX + 1 (K ∈ R) and the ellipse x ^ 2 / 5 + y ^ 2 / T = 1 with focus on the x-axis have a common point, then the value range of T is?
I don't know which one is right. Help me. I know how to work out the answer without the process
[1-5k^2,5)
[1,5)
[1,√5)
In fact, this problem needs some skills: because the straight line passes through the point (0,1), it only needs the point (0,1) on or in the ellipse
Five
Given that 12 + √ 5 = x + y, where x is an integer and 0 ＜ y ＜ 1, find the square root of X-Y + √ 5
Due to: 2
∵2＜√5＜3
∴14＜12+√5＜15
∴ X=14
∴Y=12+√5-14=√5-2
∴X-Y+√5=16
Its square root is ± 4
If the line y = KX + 1 (K ∈ R) and the ellipse x ^ 2 / 5 + y ^ 2 / T = 1 with the focus on the X axis always have a common point, then the value range of T is the same——
1: The focus is on X, which means T < 5
2: If two equations are established simultaneously and Y is eliminated, then there will be a system of linear equations with respect to x containing T. because there is a constant common point, so △ = B ^ 2-4ab ≥ 0, the range of t can be obtained
3: 1 ≤ T < 5 can be obtained by synthesizing 1 and 2
A1377051 positive solution 1 ≤ T < 5
(0,1) should be in the ellipse. 1≤t＜5
What is y equal to? Given the square root of y = X-2 + 2-x + 3, find the square root of X of Y
It is known that y = (X-2) square root + (2-x) square root + 3?
To make the radical meaningful:
x-2>=0,x>=2;
2-x>=0,x
This topic is a bit false, the topic is not clear, there are many results
If the line y = KX + 1 (K ∈ R) and the ellipse X25 + y2m = 1 always have a common point, then the value range of M is ()
A. [1，5）∪（5，+∞）B. （0，5）C. [1，+∞）D. （1，5）
When y = KX + 1x25 + y2m = 1 and Y is eliminated, we get that (M + 5k2) x2 + 10kx + 5-5m = 0, (m ＞ 0, m ≠ 5) ∵ the line y = KX + 1 (K ∈ R) and the ellipse X25 + y2m = 1 have the same common point, namely 100k2-20 (1-m) (M + 5k2) ≥ 0, M2 + 5mk2-m ≥ 0, ∵ m ＞ 0, ∵ m ≥ - 5k2 + 1, ∵
The square root of X + 1 plus the square root of Y-X equals 0. Find the square root of 2XY + 14
√(X+1)+√(y-x)=0,
If the sum of two nonnegative terms is 0, then both are 0
∴x+1=0,y-x=0
∴x=y=-1
∴√(2xy+14)=√16=4
If the line y = KX + 1 (K ∈ R) and the ellipse X25 + y2m = 1 always have a common point, then the value range of M is ()
A. [1，5）∪（5，+∞）B. （0，5）C. [1，+∞）D. （1，5）
When y = KX + 1x25 + y2m = 1 and Y is eliminated, we get that (M + 5k2) x2 + 10kx + 5-5m = 0, (m ＞ 0, m ≠ 5) ∵ the line y = KX + 1 (K ∈ R) and the ellipse X25 + y2m = 1 have the same common point, namely 100k2-20 (1-m) (M + 5k2) ≥ 0, M2 + 5mk2-m ≥ 0, ∵ m ＞ 0, ∵ m ≥ - 5k2 + 1, ∵
The function y = log takes a as the logarithm of base X. when x > 2, it always has | y | > 1?
y=log(a)x
When a > 1
log(a)2＞1
1＜a＜2
0＜a＜1
log(a)2＞-1
0＜a＜1／2
1 when a > 1, because x > 2, then Y > 0
|Y | > 1 can be written as Y > 1
So log ^ x > log ^ A
So x > A, so AA > 1
2 when 1 > a > 0, so x > 2, then Y1
It can be written as Y0, so x > A ^ - 1
So A0
1 when a > 1, because x > 2, then Y > 0
|Y | > 1 can be written as Y > 1
So log ^ x > log ^ A
So x > A, so AA > 1
2 when 1 > a > 0, so x > 2, then Y1
It can be written as Y0, so x > A ^ - 1
So A1 because x > 2 then Y > 0, so
|Y | > 1 can be written as Y > 1
So log ^ x > log ^ A
So x > A, so AA > 1
2 when 1 > a > 0, so x > 2, then Y1
It can be written as Y0, so x > A ^ - 1
So A0 y = log (a) x
When a > 1
log(a)2＞1
1＜a＜2
0＜a＜1
log(a)2＞-1
0 ＜ a ＜ 1 / 2

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