Find the maximum and minimum values of the square + root sign 2x-1 of the function f (x) = x on [1,5] I don't know what to do with the radical number. Teach me,

Find the maximum and minimum values of the square + root sign 2x-1 of the function f (x) = x on [1,5] I don't know what to do with the radical number. Teach me,

When x ∈ [1,5], f (x) is an increasing function, when x = 1, Fmin (x) = 2, when x = 5, Fmax (x) = 28
Given that vectors a and B are perpendicular to each other, vectors 3A + 2B and λ A-B are perpendicular to each other, and | a | = 2, | B | = 3. Find the value of real number λ
There are two conditions for them to be zero
2. The square of a vector is equal to the square of its module
From the condition, we know AB = 0
(3a+2b)(λa-b)=0
3λa²-3ab+2λab-2b²=0
3λ|a|²-2|b|²=0
12λ-18=0,λ=3/2
∵ A-B and a are perpendicular to each other ∵ (a-b) * a = 0 (1-2,1-x) * (1,1) = 0 - 1 * 1 + (1-x) * 1 = 0 - 1 + 1-x = 0, x = 0, x = 0, x = 0
3/4
Given that the function f (x) = root (1 + 2 ^ x + 3 ^ x * a) has a definition on (negative infinity, 1), find the value range of A
When a > = 0, 1 + 2 ^ x + 3 ^ x * a > 0
If a = 0, then G (1) > = 0
0>a>=-1
It is proved that 0 > a > = - 1 is true
g(x)=1+2^x+3^x*a >= 1+2^x-3^x=h(x)
h'(x)=2^x*ln2-3^x*ln3
When 0
How to find the range of y equal to 4x minus 5x minus 1?
y=(5x-1)/(4x-2)
=5/4(x-1/5)/(x-1/4)
=5/4(x-1/4+1/20)/(x-1/4)
=5/4（1+1/20)/(x-1/4)
=5/4+(1/16)/(x-1/4)
Domain x ≠ 1 / 4
When x ∈ (- ∞, 1 / 4), it is found that when x ∈
-∞＜x-1/4＜0
-∞＜y＜5/4
X ∈ (1 / 4, + ∞)
-∞＜x-1/4＜0
5/4＜y＜+∞
Range (- ∞, 5 / 4), (5 / 4, + ∞)
By decomposition, y = (5x-1) / (4x-2) = 5 / 4 + 1 / 4 (4x-1)
Because the value range of 1 / 4 (4x-1) is not equal to 0, then the value range of Y is y ≠ 5 / 4
X2 + Y2 + 2xy-2x-2y + 1 = 0 factorization
x2+y2+2xy-2x-2y+1=0
(x+y)²-2(x+y)+1=0
(x+y-1)²=0
x²+y²+2xy-2x-2y+1=0
(x+y)²-2(x+y)+1=0
(x+y-1)²=0
x+y=1
﹙x＋y－1﹚²＝0
(x+y)²-2(x+y)+1=0
(x+y-1)(x+y-1)=0
x+y-1=0
x+y=1
x2+y2+2xy-2x-2y+1=0
(x+y)²-2(x+y)+1=0
(x+y-1)²=0
x+y-1=0
x+y=1
Given the quadratic function y = - X & # 178; + 2x-2, if t ≤ x ≤ T + 1, t ∈ R, find the maximum value of the function
Given the quadratic function y = - X & # 178; + 2x-2, if t ≤ x ≤ T + 1, t ∈ R, find the maximum value of the function
Y = - (x-1) 2 -- 1 if t + 1
Y = - (x + 1) 2-3ruo-2
Given the vector | a | = 2, | B | = 3, a is perpendicular to B, and 3a + 2B and the input A-B are also perpendicular to each other, find the value of the real number input
λ=1.5
(3a+2b)(λa-b)
=3λa-2b+(2λ-3)ab
=0
ab=0
So λ = 2-3
λ=1.5
2/3
AB = 0 (3a + 2b) times A-B = 0
2 minus 6 points x + 4 = 3 points 2x-1 help~
2 minus 6 points x + 4 = 3 points 2x-1
Multiply both sides by six at the same time, and you get
12-（x+4）=2(2x-1)
12-x-4=4x-2
8-x=4x-2
4x+x=8+2
5x=10
X=2
Transfer the term, get 2x of 3 + X of 6 = 2 + 4 + 1
5x / 6 = 7
x=42/5
Factorization (x2-y2) - (X-Y) 2
fast
solution
(x²-y²)-(x-y)²
=(x-y)(x+y)-(x-y)²
=(x-y)[(x+y)-(x-y)]
=(x-y)×2y
=2y(x-y)
=(x+y)(x-y)-(x-y)^2=(x-y)(x+y-x+y)=2y(x-y)
Two is square
The original formula = (x + y) (X-Y) - (X-Y) & #178;
=（x-y）【（x+y）-（x-y）】
=2y（x-y）
（x²-y²）-（x-y）²
=(x-y)(x+y)-(x-y)²
=(x-y)[x+y-(x-y)]
=(x-y)(x+y-x+y)
=2y(x-y)
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Find the maximum value of function f (x) = - X & # 178; + 2aX + 2, X ∈ [- 5,5]~
f(x)=-x²+2ax+2
=-(x-a)²+a²+2
It can be obtained that x = a is axisymmetric
When - 5 ≤ a ≤ 5, when x = a, there is a maximum value of: A & # 178; + 2
When A5 and x = 5, the maximum value is 10a-23
-b/2a = -2a/2 = -a
If a = 0, then f (max) = f (5) = f (- 5) = 27
If a > 0 then - A