It is known that m (x, y) is a set of inequalities where x is greater than or equal to 0, less than or equal to root 2, y is less than or equal to 2, and X is less than or equal to root sign 2Y If point a (root 2,1), then z = vector OM, the maximum value of point multiplication vector OA is

It is known that m (x, y) is a set of inequalities where x is greater than or equal to 0, less than or equal to root 2, y is less than or equal to 2, and X is less than or equal to root sign 2Y If point a (root 2,1), then z = vector OM, the maximum value of point multiplication vector OA is

｛0≤x≤√2
 {y≤2
OM·OA
=(x,y)·(√2,1)
=√2x+y
The optimal solution corresponding to the maximum value is (√ 2,2)
The maximum value of OM · OA is
√2*√2+2=4

If y √ 2 √ is given, then the maximum value of Y √ 2 x ≤ 2 x is given

Draw a picture, obviously. When x = radical 2, y = 2, the maximum value is 4

It is known that the inequality system {0 ≤ x ≤√ 2, y ≤ 2, X ≤√ 2Y} in the plane rectangular coordinate system xoy is known. If M (x, y) is a moving point on D and a (√ 2,1), then 8 = the maximum value of vector OM and vector OA

Inequality system {0 ≤ x ≤√ 2, y ≤ 2, X ≤√ 2Y}
The area is trapezoidal oabc, as shown in the figure
A(√2,1),B(√2,2),C(0,2)
z=OM·OA
=(x,y)·(√2,1)
=√2x+y
Let z = 0, make a line of √ 2x + y = 0,
It can be seen that the optimal solution of Z Max is B (√ 2,2)
zmax=√2*√2+2=4
The maximum value of OM · OA is 4

Given x ^ 2 + y ^ 2 = 1, x, Y > 0, prove: x + 2Y ≥ root 5, please use the knowledge of basic inequality to prove Wrong, should be ≤ root 5 It has been said that the proof of basic inequality knowledge is the related knowledge of a + B ≥ 2 root sign ab

According to the basic inequality a 2 + B 2 ≥ 2 AB, we can get: (2 x) 2 + y 2 ≥ 2.2 x · y = 4 XY

Basic inequality 2Y / x + X / Y > = 2 radical 2 When 2Y / x = x / y, the equal sign holds. How to find x = radical 2 - 1, y = 1 - (radical 2 / 2)

2y/x+x/y≥2√2
(2y^2+x^2)/(xy)≥2√2
[(2y^2+x^2)/(xy)]^2≥(2√2)^2
[4y^4+4(x^2)(y^2)+x^4]/[(x^2)(y^2)]≥8
4y^4+4(x^2)(y^2)+x^4≥8(x^2)(y^2)
4y^4+4(x^2)(y^2)+x^4-8(x^2)(y^2)≥0
4y^4-4(x^2)(y^2)+x^4≥0
(2y^2-x^2)^2≥0
It can be seen that when x ^ 2 = 2Y ^ 2, the equal sign holds
That is: X / y = 2Y / X
The exact values of X and y cannot be obtained

Let y = −x2+2x+3− The image of 3 (x ∈ [0,2]) rotates anticlockwise about the origin of coordinates (θ is an acute angle). If the curve obtained is still a function image, then the maximum value of θ is______ .

Let f (x) = − x2 + 2x + 3 − 3, according to the monotonicity of quadratic function, we can get that the function is increasing function on [0,1] and decreasing function on [1,2]. If the tangent slope of function is k at x = 0, then k = f '(0) ∵ f' (x) = 12 · (− x2 + 2x) ′ − x2 + 2x + 3 = − x + 1 − x2 + 2x +

Let f (x) = log root 2 (x + a) be known to cross the origin (1) Find the value of A (2) If 2F (2-1 under the radical) = f (x-3) + F (x-4), find the value of X

I: the function can be reduced to f (x) = 2log2 (x + a);
∵f(0)=0 ;
(0 + a) = 1; push out a = 1;
Ⅱ: F (x) = 2log2 (x + 1) is known from I;
So the above inequality is
4log2 (root 2) = 2log2 (X-2) + 2log2 (x-3)
2log2(2)=2log2[(x-2)(x-3)]
So (X-2) (x-3) = 2
(x-1)(x-4)=0
X = 1 or x = 4
Because the domain of F (x) is x > 0
So x-3 + 1 > 0 and x-4 + 1 > 0
Therefore, the domain is defined as x > 3 in the inequality
X = 1 (minus)
In conclusion, x = 4;

On the image of the inverse scaling function y = 6 / x, the point whose distance from the origin o of the coordinate is equal to the root 13 is

Make a set of equations
Square of X + square of y = 13
Y=6/X
There are four points
(2,3)
(-2,-3)
(3,2)
(-3,-2)

Let f (x) = 2 | x + 1 | - | X-1 | and make f (x) ≥ 2 The value range of X of 2

Since y = 2x is an increasing function, f (x) ≥ 2
2 is equivalent to | x + 1 | - | X-1 | ≥ 3
2，①
(1) When x ≥ 1, | x + 1 | - | X-1 | = 2, then the formula ① will always hold,
(2) When - 1 ＜ x ＜ 1, | x + 1 | - | X-1 | = 2x, and ① is expressed as 2x ≥ 3
2, namely 3
4≤x＜1，
(3) When x ≤ - 1, | x + 1 | - | X-1 | = - 2, there is no solution to equation 1
To sum up, the value range of X is [3
4，+∞）．

In the function y equal to 1 / 2 of the root sign x, the value range of the independent variable x is?

x> 2 is the root sign, X-2 is greater than or equal to 0, then x is greater than or equal to 2; and because the denominator is not 0, that is, X is not equal to 2, so x > 2
Ask, you are also the third year of junior high school/